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THE CALCULATIONS 

OF 

GENERAL CHEMISTRY 


WITH 

DEFINITIONS, EXPLANATIONS, AND 
PROBLEMS 


BY 

WILLIAM J. HALE, Ph.D. 

ASSISTANT PROFESSOR OF CHEMISTRY IN THE UNIVERSITY 
OF MICHIGAN 


SECOND EDITION , REVISED 



NEW YORR 

D. van nostrand company 

23 MURRAY AND 27 WARREN STS. 

1910 



Copyright, 1910, 

BY 

D. VAN NOSTRAND COMPANY 
NEW YORK 



Stanbope fl>ress 

F. H. GILSON COMPANY 
BOSTON. U.S.A. 




£ Cl. A280772 


To the Memory of 

Irenrg IBarhpr 

LATE PROFESSOR OF ORGANIC CHEMISTRY IN 
HARVARD UNIVERSITY 


IN APPRECIATION OF HIS MOST INSPIRING INFLUENCE 

THIS BOOK 

IS GRATEFULLY DEDICATED 


# 


PREFACE TO THE FIRST EDITION 


The justification for a book of this character is to be 
found in the realization that many of the mathematical 
applications of our fundamental conceptions in chemistry, 
even upon the most elementary points, remain uncompre¬ 
hended by students several years advanced in the study of 
the science. The great attention given to the study of 
Physical Chemistry, and the far-reaching importance 
attached to the interpretation of chemical phenomena in 
the light of modern theories, make it absolutely essential 
that more time be given in the college courses on General 
Chemistry to these mathematical demonstrations. 

Unless a clear and concise exposition of the methods of 
calculation is presented in the very first course of college 
work, the progress of the student is greatly hindered. On 
the other hand, too great a mass of mathematical data 
may impede rather than promote this progress. The aim, 
therefore, has been to limit these calculations to those 
subjects generally regarded as fundamental, and in which 
the student should receive the most complete and thorough 
drill. Such may rightfully constitute our “Arithmetic of 
Chemistry.” Of the many important subjects omitted, 
the greater number possess a theoretical bearing which 
brings them more properly into those courses following 
General and Analytical Chemistry or into the study of 
Physical Chemistry itself. 

The absence of mathematical training as a basis for the 
study of chemistry constitutes a widely prevailing defect 
in the education of chemists at the present day. The 
vii 



PREFACE 


yiii 

appearance of works on General Chemistry from the stand¬ 
point of Physical Chemistry, with their extensive adoption 
in our American colleges and universities, is strongly indic¬ 
ative of the mathematical trend in modern chemical 
instruction. Under this influence the presentation of the 
methods of chemical calculations in their simplest possible 
form should facilitate and extend the use of these modern 
texts and at the same time operate favorably in the eradi¬ 
cation of former defects. Primarily, however, this ele¬ 
mentary presentation is intended to accompany the 
laboratory work in General Chemistry and to help the 
student to a better understanding of the many possible 
conditions that may arise in the course of experimental 
work. The ideal plan, therefore, should be to associate 
these problems and their solutions with the actual labora¬ 
tory practice. In the study of the last two, and possibly 
three, chapters only a very few correlated experiments 
may be found to fit properly into the laboratory courses 
on General Chemistry; but the important bearing of the 
topics here outlined upon the thorough grounding of the 
student’s ideas of chemical reactions will make it impera¬ 
tive that a number of the more simple illustrations be 
given by way of lecture experiments. Especially is this 
true of those reactions relating to combinations between 
gases by volume. 

For the greater part of material presented no originality 
is claimed, but gratitude is freely expressed to the authors 
of the many valuable and admirable treatises on Stoi¬ 
chiometry. In the manner of presentation a somewhat 
different plan has been followed from that usually found 
in books of this nature. This consists in a gradual intro¬ 
duction of each new condition properly falling under the 
consideration of some one subject, and the final develop¬ 
ment of the subject in its entirety from all the conditions 


PREFACE 


IX 


thus considered. By this method it is thought to obviate 
the dogmatic manner of presenting mathematical facts 
and to make the student realize immediately the connec¬ 
tion between all of our fundamental laws of energy and 
matter. With this train of thought connecting each and 
every point to that preceding, it is believed that the 
student will gain a better and firmer hold upon the entire 
subject-matter rather than upon the isolated points. 

This book in itself comprises no more than is presented 
by the author to students in their first year of chemistry 
at the university. The problems at the end of each chap¬ 
ter serve as a guide and basis of selection for examinations 
throughout the course. Great emphasis, in fact, is laid 
upon this side of the student's development as the highest 
and most beneficial for his future advancement. 

In general the field of use for a book of this type will be 
among colleges and universities. It is, however, sincerely 
hoped that at least the first portion of the book may have 
a helpful and profitable bearing upon the instruction of 
chemistry in the high schools. 

For the many valuable suggestions and the interest with 
which criticism has been freely extended in the revision of 
the manuscript, the author expresses his sincere apprecia¬ 
tion and lasting indebtedness to Professor Charles Loring 
Jackson of Harvard University, Professor Alexander Smith 
of the University of Chicago, and Professor S. Lawrence 
Bigelow of the University of Michigan. 

WILLIAM J. HALE. 


Ann Arbor, May, 1909. 


PREFACE TO THE SECOND EDITION 


With a second and larger edition it has seemed best to 
make all possible minor changes in both text and problems. 
The work of revision has been greatly facilitated through 
the kind suggestions of those who have found the book help¬ 
ful, and of others who have shown especial interest in 
this exposition of the subject. For all of which grateful 
appreciation is heartily expressed. 

THE AUTHOR 

Ann Arbor, January, 1910. 


x 



CONTENTS, 


Chapter Page 

I. Units op Measurement. 1 

II. Density and Specific Gravity . 5 

III. The Effect of Pressure upon Gases. 9 

IV. The Effect of Temperature upon Gases. 15 

V. The Combined Effect of Pressure and Tem¬ 
perature upon Gases. Partial Pressures .. 21 

VI. Avogadro’s Hypothesis and Some of Its Ap¬ 
plications . 33 

VII. The Law of Definite Proportions. 46 

VIII. The Derivation of Chemical Formula. 59 

IX. Calculations Depending upon Chemical Equa¬ 
tions . 82 

X. Normal Solutions.. 106 

XI. Combinations between Gases by Volume. 122 

XII. Complex Equations. 148 

Appendix. 167 

Index. 173 


xi 



















♦ 







% 


♦ 












CHEMICAL CALCULATIONS 


CHAPTER I. 

UNITS OF MEASUREMENT. 

The fundamental units employed in chemical calcula¬ 
tions will be defined at the outset to insure their clear and 
consistent use. 

Time. — As the unit of time we have adopted the 
second, g^*^ part of the mean solar day, or that period 
elapsing between the successive daily transits of the sun 
across a meridian. 

Distance or Length. — The meter as the standard of 
length is fixed as that distance between two marks on a 
platinum-iridium bar preserved in Paris when this bar is 
at the temperature of 0° C. It is approximately one 
forty-millionth of a meridian, or one ten-millionth of the 
earth's quadrant from pole to equator. Larger values 
in multiples of the meter by ten, one hundred and one 
thousand are named by use of the Greek prefixes, — deca¬ 
meter, hectometer and kilometer (km.), — while submul¬ 
tiples by the same values receive names from the Latin 
prefixes, — decimeter (dcm.), centimeter (cm.) and milli¬ 
meter (mm.). The one one-hundredth part of the meter, 
the centimeter, is usually defined as the unit of length. 
As still smaller values we employ the micron (/*), the one 
one-thousandth of a millimeter, and the millimicron (///*), 
the one one-thousandth of a micron. 

1 



2 


CHEMICAL CALCULATIONS 


Volume and Mass. — The cubic decimeter or liter is 
the standard of volume. The standard of mass is the 
kilogram, or a mass of platinum-iridium in block form, 
preserved in Paris, and originally intended to have the 
same mass as a cubic decimeter of water at its greatest 
density, 4° C. It is, however, slightly less than a cubic 
decimeter of water which at 4° C. weighs 1.000013 kilo¬ 
grams. Both the liter and kilogram are somewhat large 
for general scientific purposes. It is customary, therefore, 
to use the one one-thousandth part of each, — the cubic 
centimeter (cm. 3 or c.c.) as the unit of volume and the 
gram (g.) as the unit of mass. The mass of 1 c.c. of water 
at 4° C. is considered as 1 gram. The slight discrepancy 
between this value and the true one need be considered 
only in the most exact calculations. 

Through a combination of the fundamental units just 
mentioned we arrive at a standard system in terms of 
which so many of our important units of measurement 
may be defined. This system of units, known as the 
centimeter-gram-second (C.G.S.) system, has met with 
universal adoption throughout the scientific world. Its 
applications may be noted in the following paragraph. 

Force and Energy. — When a body moves at a uniform 
rate through a unit of distance, the centimeter, in a unit 
of time, the second, it is said to have a unit of velocity 
(abbreviated cm./sec.). When the change in velocity 
during one second is one centimeter per second we have 
a unit of acceleration. That force which will give to unit 
mass unit acceleration is called the unit of force or dyne. 
Now the same force operating upon different bodies does 
not produce in each the same acceleration. When, how¬ 
ever, the same acceleration is attained in the several 
cases the bodies must have equal masses. The work 
done by the force of one djme in producing a displace- 


UNITS OF MEASUKEMENT 


3 


ment in its own direction of one centimeter is called the 
unit of work or erg . When a body gains or loses energy 
the amount of energy may be measured in units of work, 
and when the' work done upon a body imparts to it a 
; certain velocity, the body, by virtue of this, is said to 
possess kinetic energy. The amount of energy so pos- 

MV 2 

sessed may be expressed by the formula, E = , in 

which E represents the energy, M the mass and V the 
velocity. Mass, then, is in itself but a “ measure of the 
kinetic energy which a body possesses when it has a 
definite velocity.” — Ostwald. 

Standards of Temperature and Pressure. 

The accuracy with which measurements are conducted 
requires certain definite and constant conditions of tem¬ 
perature and pressure, the effect of changes in which will 
be studied in a later chapter. The standards adopted with 
these factors will receive only brief mention at this 
point. 

Temperature. — The freezing-point of pure water is 
taken as the normal temperature and made 0° upon the 
Celsius or centigrade thermometer. The boiling-point of 
pure water at 760 mm. pressure is registered at 100° on 
this scale, and the intervening range of temperature 
between the freezing- and boiling-points of water is di¬ 
vided into one hundred equal parts or degrees centigrade. 
As pressure exerts a considerable influence upon the 
boiling-points of liquids, all temperatures are referred to 
the normal condition of pressure. 

Pressure. — The standard condition of pressure, the 
normal pressure, is taken as that pressure which the 
atmosphere exerts at sea level in a latitude of 45°. This 




4 


CHEMICAL CALCULATIONS 


pressure is sufficient to sustain a column of mercury, at 
0° C., 76 cm. (760 mm.) in height. In a column of this ! 
height and 1 sq. cm. cross section there are exactly 76 c.c. 
of mercury, and since, volume for volume, mercury is 
13.596 times as heavy as water and 1 c.c. of it therefore 
weighs this number of grams, we have at once the value ] 
76 X 13.596 = 1033.2 grams as the weight of the atmos- i 
phere per square centimeter. By reference to a barometer 
the actual weight of the atmosphere in millimeters of 
mercury is observed under the various conditions. These 
barometric readings are usually made at other tempera¬ 
tures than 0° C.; the correct readings, therefore, for milli¬ 
meters of mercury at 0° C. must be calculated by use of a 
proper table showing the expansion of mercury with tem¬ 
perature (Appendix I). Gases measured in vessels in¬ 
verted over a liquid will of course have the same pressure! 
as the atmosphere (recorded by the barometer) when the 
levels of the liquid in the vessel and outside of it are equal. 
The temperature of the gases and liquid should of course 
be alike. 




CHAPTER II. 


DENSITY AND SPECIFIC GRAVITY. 

The term density is used to designate the mass in unit 
of volume. As this is a constant for any substance under 
given conditions of temperature and pressure, we may call 
it the absolute density . It is expressed simply as D = M/V, 
where D, M and V represent respectively density, mass 
and volume. When the absolute density of one substance 
is referred to that of another, i.e., a relation between these 
two densities determined upon the basis of one of them as 
a standard, we have a value known as the specific or rela¬ 
tive density, or what has been more commonly called the 
specific gravity. The relative density of a substance is not 
affected by the force of gravity, since any change in this 
force will operate equally upon the absolute densities of 
all substances as well as upon the standard. ^"The relative 
density or specific gravity may be defined simply as the 
ratio between the weight of any substance and that of an 
equal volume of the standard substance when this latter 
is considered as unity. As a standard for the specific 
gravity of liquids and solids, the most natural choice is 
distilled water, which is considered at the point of its 
greatest density, namely 4° C. Determinations made at 
other than this temperature ( e.g . 15°) are usually referred 
to the density of water at 4° C., and expressed thus: 
sp. gr. at 15°/4°. 

In the case of gases, the values obtained for the absolute 
densities, the number of units of mass per unit of volume, 
are exceedingly small. By reason of this it is more cus- 

5 



6 


CHEMICAL CALCULATIONS 


tomary to multiply these values by 1000 or, what is the 
same thing, to determine the number of units of mass in 
1 liter instead of 1 c.c. Though the densities of gases 
referred to some standard such as air are known as relative 
densities , the term vapor density meets with constant use 
when the relative density is determined for a liquid or 
solid in the state of vapor. Since the composition of the 
atmosphere is somewhat variable, the relative densities 
upon this standard cannot possess much accuracy. The 
employment of a pure gaseous substance, such as hydrogen, 
gives much better results. As the lightest gas, all other 
densities referred to it assume values greater than unity. 
The absolute density of hydrogen, the weight in grams of 
1 liter at 0° and 760 mm., is 0.08987. The weight of an 
equal volume, 1 liter, of oxygen under the same conditions 
is 1.429 grams; hence with reference to hydrogen as unity, 
oxygen will be found to have a relative density of 15.90, 
i.e.j 1.429 is 15.90 times 0.08987, or, by simple proportion: 

0.08987: 1.429 = 1: x, or x = 15.90. 

The Oxygen Standard. — The intimate relation (to be 
noted later) between the molecular weight of a substance 
and the volume which this weight occupies in the state of 
vapor has led us to refer all densities of gaseous substances 
to the density of that substance which is to serve as the 
basis for molecular weights. Formerly hydrogen, as the 
lightest substance, served this purpose, and consequently 
the close relationship between densities and molecular 
weights was apparent. 

In recent times oxygen, with the value 32, has been 
adopted as the basis of molecular weights by reason of 
the great importance of this element in its numerous 
combinations with other elements and for reasons that 
will be made clear after further considerations. The 


DENSITY AND SPECIFIC GKAVITY 


7 


absolute density of oxygen, therefore, is now taken as the 
standard of gas densities and made equal to unity. Upon 
this basis the close relationship between the densities of 
gases and their corresponding molecular weights will be 
as well attained as originally, when hydrogen was the 
standard. The relative densities of a number of sub¬ 
stances, however, will fall below unity. Thus the relative 
density of hydrogen becomes on this scale 0.0629, a value 
easily derived from the simple proportion: 

1.429 : 0.08987 = 1: x. 

In order to avoid these small values some chemists prefer 
to assign to oxygen the absolute density of 16 instead of 
unity. From the relations already noted between oxygen 
on the one hand and hydrogen, as the lightest substance, 
on the other, we readily see that the relative densities of 
all other substances are thus brought to values greater 
than unity. By the adoption of 32 instead of 16, as this 
basis of densities, the relative densities may be made to 
coincide at once with the molecular weights.* 

Under all conditions the measure of the volumes of 
gases is conducted at definite temperatures and pressures. 
The influence* of change in these factors will be discussed 
in the succeeding chapters. Whatever the volume of a 
gas may be, its relative density will be determined from 
the ratio of the observed weight per volume to the weight 

* Strictly speaking, an exact agreement on this basis between 
relative densities and molecular weights is found true only in the 
case of perfect gases (c/. footnote, p. 34). In the study of actual 
weights of equal volumes of gases it seems preferable, therefore, 
to base the comparison upon the weight of the standard as unity. 
The significance of the theoretical or calculated values, upon the 
basis of 32 as the chemical unit, will be noted in succeeding 
chapters. 




8 


CHEMICAL CALCULATIONS 


of an equal volume of oxygen under like conditions, when 
this ratio is brought over to the basis of oxygen as unity. 

Example 1 .— The actual weight of 1 liter of carbon dioxide, at 
standard conditions, is 1.977 grams. One liter of oxygen weighs 

1.429 grams. What is the relative density of carbon dioxide? 

The ratio between these two densities is now to be 

referred to the density of oxygen as unity. That is, the 

1.429 . 1 

ratio ^ is to be made equal to the ratio -, where x is 

the relative density of carbon dioxide. We thus have 

1.429 1 . , 

- Q77 = -, or, as more commonly expressed, 

l.y* i x 


1.429 : 1.977 = 1:*. 

From which x is found to be 1.384. 

PROBLEMS. 

1. A piece of metal weighing 30 grams displaced 20 c.c. of 

water (i.e. f its own volume). What is the relative density of 
this metal referred to water? Ans . 1.5. 

2. A vessel weighing 6.448 grams weighed 7.963 grams when 
filled with water and 8.266 grams when filled with a salt solu¬ 
tion. What is the relative density of this solution referred to 
water? (The temperature constant throughout.) 

Ans. 1.2. 

3. The weight of 1 liter of aqueous vapor at 0° and 760 mm. 
is 0.8045 gram. What is its relative density? .4ns. 0.5630. 

4. Calculate the relative density of mercury vapor, 1 liter of 

which at standard conditions (0° and 760 mm.) weighs 8.87 
grams. Ans. 6.207. 

5. Calculate the relative density of hydrogen chloride, 5 liters 
of which under standard conditions weigh 8.205 grams. 

Ans. 1.148. 

6. Calculate the relative density of chlorine, 100 c.c. of which 
at standard conditions weigh 0.322 gram. Ans. 2.253. 




CHAPTER III. 


THE EFFECT OF PRESSURE UPON GASES. 

THE LAW OF BOYLE. 

The Relation of Volume to Pressure. — When a gas is 
subjected to increase of pressure the volume decreases. 
On release of the pressure the volume increases, regain¬ 
ing its former volume only at the original pressure. The 
Law of Boyle states this as follows: At constant tem¬ 
perature the product of the volume of a gas (V) by its 
pressure (P) is a constant, or P X V = K. Thus a 
volume of gas (1) at a pressure of one atmosphere (1) 
will be reduced when under a pressure of two atmos¬ 
pheres to one-half of its original volume, or, in the equa¬ 
tion P X V = K, the values 1X1 = 1 will become 
2 X \ = 1. This constancy, however, applies to all 
gases, and hence the product PV in one case will be equal 
to the product P'V' for any other case when either P or V 
or both change, the temperature considered constant 
throughout; hence the equation PV = P'V'. 

In dividing this equation, PV = P'V', through by the 
quantity PV' and canceling like terms in numerator and 
PV P'V' V P' 

denominator, we obtain = pyT or y 7 = p" ’ ^ wo 

fractional quantities may be expressed also by the propor¬ 
tion V : V' = P' : P. This is exactly what is indicated 
above, where a volume of gas under a definite pressure 
will have its volume halved under double this pressure, 
etc., or at constant temperature the volume of a gas is 
inversely proportional to its pressure (a -general form for 

9 


10 


CHEMICAL CALCULATIONS 


the statement of Boyle’s Law). Strictly speaking, the 
Law of Boyle holds true only for a perfect gas. For prac¬ 
tical purposes the law gives a sufficiently accurate means 
of studying volume and pressure relations. 

In order to ascertain what a given volume of gas may 
become at some new pressure, the temperature constant 
throughout, we have only to apply the formula PV = P'V', 
in which we may designate the new values by the prime 
marks. 

Example 2 .—200 c.c. of hydrogen measured at 750 mm. 
pressure will occupy what volume at the standard pressure 
(760 mm.), temperature a constant? 

PY = P'V', or V' = V (P/P'). Substituting here the 
respective values P' = 760, P = 750 and V = 200, we 
obtain the expression V' = 200 X 750/760, which is easily 
solved to a value of 197.3. 

In general terms, we shall say that a change in the volume 
of any gas by change in pressure is readily calculated from 
the values given, by multiplying the volume by the fraction 
formed from the original pressure as numerator and new 
pressure as denominator. If this new pressure is greater 
than the original, then the fraction will have a value less 
than unity and the new volume (the product of fraction 
by original volume) will be proportionately less; con¬ 
versely, if the new pressure is smaller than the original 
pressure, the fraction (formed from the pressures) be¬ 
comes larger than unity in value, and the new volume 
must increase over that of the original. It comes, there¬ 
fore, to the same end if one ask himself the question: 
u Is the new pressure greater or less than the original? ” 
If greater, then the fraction formed from the pressures 
must be so arranged that the new volume of the gas will 
be smaller, i.e., the fraction must have as its numerator 


THE EFFECT OF PRESSURE UPON GASES 11 

the smaller of the two pressure values, and as its denomi¬ 
nator the larger. This fraction less than unity in value 
will, as was just shown, reduce the volume which it mul¬ 
tiplies in the correct proportion. If the new pressure is 
smaller than the original, the diminution in pressure must 
be accompanied by an expansion in volume, and, accord¬ 
ingly, the fraction formed from the pressures, which we 
may conveniently call the pressure-fraction , takes the 
larger value in the numerator and smaller value in the 
denominator. From the product of this fraction, greater 
than unity, by the original volume we obtain the new and 
correspondingly increased volume. 

The Relation of Pressure to Volume. — Just as change 
in pressure produces a change in volume, so also a change 
in volume will produce a change in the pressure of a gas. 
The effect of this change in volume upon the pressure is 
readily determined from the equation PV = P'V', and 
takes the form of expression: 

P' = P (V/V'). 

As the new volume V' increases or decreases with reference 
to the original, the volume-fraction becomes respectively 
less or greater than unity in value. 

Examples .—A volume of gas measuring 100 c.c. at 750 
mm. was expanded to 500 c.c. at constant temperature. What 
was the pressure of the gas at this final volume? 

In order to arrange the fraction (formed from the 
volumes) in such a way as to give a value less than unity 
and one that will reduce the given pressure in the ratio 
indicated by Boyle’s Law, it is necessary here to place the 
smaller number in the numerator. The product of this 
volume-fraction by the original pressure P, or 750 mm., 
gives the new pressure sought: 750 X 100/500 = 150. 


12 


CHEMICAL CALCULATIONS 


The Relation of Density to Pressure. — The absolute 

density of a gas, the mass per liter, may be seen now to 
vary with the pressure to which the gas is subjected. 
With reference to the formula D = M/V, we are aware 
that the weight or mass is unaffected by any changes 
in temperature or pressure. The volume (V), however, is 
altered by either or both of these influences. Thus with 
an increase of pressure it decreases and hence the absolute 
density must increase. In other words, both the absolute 
density and the pressure of a definite quantity of gas vary 
inversely with the volume when the temperature is con¬ 
stant, and hence are directly proportional to each other. 
Therefore, in the equation PV = P'V' we may substitute 
D for P throughout and obtain 

DV = D'V', or D' = D (V/V'). 

Example 4 . —100 c.c. of hydrogen, the absolute density of 
which (the weight of 1 liter) is 0.Q8987, will have what density 
when expanded to 200 c.c. in volume at a constant tem¬ 
perature? 

Here the increase in volume means a decrease in the 
pressure and hence in the density. The fraction formed 
from the volumes, the volume-fraction, must be arranged 
so as to produce this decrease corresponding to Boyle’s 
Law and, therefore, the numerator must contain the 
smaller of the two values. When this quantity 100/200. 
is multiplied by the density 0.08987, we obtain the new 
density 0.0449. 

This is identical with saying that at this new and larger 
volume the actual weight of the gas has not changed, or 
the original weight of the 100 c.c., 0.008987 gram, has 
remained a constant. The mass per volume, however, 
has been changed, in that this weight of gas is now spread 


THE EFFECT OF PRESSURE UPON GASES 13 


out over twice the original volume. The actual weight 
of 100 c.c. of the rarefied gas is just one-half of the weight 
of the total volume, 200 c.c., or 0.00449 gram. As a 
basis for calculations of this nature, that general property 
of homogeneous substances is of course understood: The 
weight of any fractional part of a volume bears the 
same ratio to the total weight as this fractional volume 
itself does to the total volume. 

It should be remembered in this connection that the 

relative density of a solid or liquid may vary with the 

temperature or pressure (p. 5), since the effect of change 

in these factors upon the volume of a substance and upon 

that of the standard chosen may not be the . same. The 

relative density of a gas, however, is constant, since 

changes in these factors affect all gases alike. 

* 

PROBLEMS. 

0 

7. 512 c.c. of hydrogen at a pressure of 744 mm. will occupy 

what volume at a pressure of 790.5 mm., the temperature 
constant? Ans. 481 c.c. 

8. 240 c.c. of gas at a pressure of 740 mm. were admitted into 
an empty vessel of 800 c.c. capacity. What was the pressure of 
the gas at this new volume, temperature constant? 

Ans. 222 mm. 

9. 500 c.c. of oxygen, absolute density 1.429, were com¬ 
pressed to a volume of 125 c.c. at constant temperature. What 
was the density of the gas at this final volume? What would 
be the weight of 50 c.c. of the compressed gas? 

Ans. 5.716, 0.2858 gram. 

10. A volume of gas measuring 600 c.c. at 760 mm. pres¬ 

sure was expanded to a volume of 1000 c.c. at constant tem¬ 
perature. What was the final pressure of the gas, and what 
fractional change in its absolute density must have followed 
this expansion? Ans. 456 mm. 

Density t of former value. 


14 


CHEMICAL CALCULATIONS 


11. 500 c.c. of a gas, the absolute density of which is 6, 
must be reduced to a density of 0.75 at constant temperature. 
What will be the volume of this rarefied gas? Calculate also 
the weight of 400 c.c. of the rarefied gas. 

Ans. 4000 c.c.; 0.3 gram. 

12. A volume of gas weighing 5 grams was expanded, at a 

constant temperature, till the pressure was reduced to one-half 
of its former value. 500 c.c. of the rarefied gas weighed 1.25 
grams. What was the original volume of the gas? Calculate 
also the original density assuming the original observations 
made at standard conditions. Ans. 1000 c.c. 

Density 5. 


CHAPTER IV. 


THE EFFECT OF TEMPERATURE UPON GASES. 

THE LAW OF CHARLES. 

With pressure constant the volume of a gas is found to 
vary with the temperature. The Law of Charles (known 
also as Gay-Lussac’s Law), states this as follows: Under 
constant pressure the volume of a gas varies directly with 
the temperature upon the absolute scale. 

The Determination of the Coefficient of Expansion. — 
The standard condition of temperature, 0° C., becomes on 
the absolute scale 273°. This is calculated from the fact 
that, given any volume of gas at 0° C., the effect of cooling 
upon this gas will bring about a diminution in its volume 
by 1/273 for each degree centigrade below the centigrade 
zero, consequently at — 273° C. the volume will have been 
diminished by 273/273 of itself, or by its entirety. This 
point, therefore, is regarded as the zero point or lowest 
point at which a gas may possibly come under considera¬ 
tion. At or below this point no gas could exist as such. 
Though this temperature has not as yet been reached, 
it follows that all gases will be liquefied above it. A 
volume of gas measuring 273 c.c. at 0° C. would lose per 
degree of cooling 1 /273 of itself, or a single unit of volume. 
At one degree of temperature above the absolute zero 
(designated as 1° A.) the gas, having cooled through 272° 
and lost 272/273 of its volume at 0° C., would consequently 
occupy but 1 c.c. At 2° A. its volume would be 2 c.c., 
having lost here 271/273 of its volume at 0° C. At 
273° A. its volume would be again 273 c.c.; in fact through- 
15 


16 


CHEMICAL CALCULATIONS 


out the entire range, 0° A. to 273° A., a unit of volume 
corresponds to a unit of temperature, and the one is there¬ 
fore directly proportional to the other, i.e., the volume 
varies directly with the temperature. Upon passing the 
point 273° A. the same principle holds. For each degree 
of temperature there is an expansion in volume corre¬ 
sponding to 1/273 of the volume at 273° A. or 0° C. At 
283° A. the volume of a gas measuring 273 c.c. at 273° A. 
will have increased by 10/273 of this volume and measure 
283 c.c. The rate of expansion above 0° C. is identical 
with that of cooling below this temperature. This ratio 
between the increase in volume per degree and the total 
volume of a gas at 0° C. —called the coefficient of expan¬ 
sion and often expressed decimally as 0.00367 in place of 
the fraction 1/273 — was determined by measuring the 
expansion of gases from 0° C. upward and this upon 
the centigrade scale; consequently the determination of 
the zero point on the absolute scale is made in centigrade 
units, and each degree of temperature centigrade must 
be equivalent to one degree on the absolute scale. If 
273° A. is the zero of centigrade, then 283° A. is equal to 
10° C. and so on. In order to preserve this direct pro¬ 
portionality between volume and temperature we have 
only to convert the centigrade temperatures into their 
corresponding values upon the absolute scale. By reason 
of the equivalence in the units we add 273° to the read¬ 
ings above 0° C. and subtract readings below 0° C. from 
273°; the results represent absolute temperatures (desig¬ 
nated by T in contradistinction to t for the centigrade 
temperatures). 

With this one volume, 273 c.c. as above chosen, the 
absolute temperature coincides numerically throughout 
with the volume in cubic centimeters, and simplifies the 
calculations to adding or subtracting single units. Had 


THE EFFECT OF TEMPERATURE TJPOH GASES 17 

any other volume of gas at 0° C. been considered, the 
same ratio between volume and temperature would be 
found—a direct proportionality—but the variations would 
not be in single units. The equation which represents 
this change in volume V under a definite change in tem¬ 
perature t (centigrade) may be simply expressed as 
V' = V (1 + at), where a is the coefficient of expansion 
per degree (1/273) and t the number of degrees centigrade 
through which it operates. Thus for a definite rise in 
temperature, t, the total expansion will be equal to that 
fractional part of the original volume as is indicated by 
the quantity at, i.e., (1/273 X t) V; and any new volume 
V 7 must be equal to the original volume Y plus the total 
expansion: 

V' = V + (1/273 X 0 Vor 
V' = V (1 + t/ 273) or 
V' = V (1 + at). 

It is regarded as more helpful in this work to arrange the 
original and new values as members of a simple propor¬ 
tion, in which, of course, only the absolute temperatures 
are considered: V' : V = T' : T. The application of the 
Law of Charles to various cases may be illustrated now 
by examples. 

The Relation of Volume to Temperature. 

Example 5. — A volume of gas measuring 200 c.c. at 20° C. 
will have what volume at the standard condition, 0° C. 

From the direct proportionality between volume and 
absolute temperature we have 

V : V' = T : T', or V' = V (T'/T), 

where T and T' represent, respectively, the original and 
new temperatures expressed in absolute units. In the 


18 


CHEMICAL CALCULATIONS 


example above, T = 20° + 273° = 293°; the new tem¬ 
perature is 273°. The change in volume of this gas must 
follow according to the proportion: 

200 : x = 293 : 273, or x = 200 (273/293), or * = 186.35 

Here we see the original volume is to be multiplied by a 
fraction formed from the two temperatures, the tempera¬ 
ture-fraction, and again we may apply the question as in 
the consideration of Boyle’s Law*: “Is the new volume to 
be smaller or greater than the original volume?” If 
greater, then the larger number (the degrees in absolute 
units) must be set over the smaller in order that the frac¬ 
tional quantity, when multiplied by the original volume, 
may increase that volume in the proportion indicated by 
these temperature values. If the new volume is to be 
smaller, then the numerator must contain the smaller of 
the two numbers and the product accordingly will be 
proportionately less. 

The Relation of Temperature to Volume. 

Example 6. — A volume of gas measuring 100 c.c. at 17° C. 
was expanded by warming to a volume of 250 c.c. at constant 
pressure. To what temperature must the gas have been heated 
to bring about this expansion? 

From the Law of Charles we draw up the equation 
T' = T (V'/V). In conformity with the direct propor¬ 
tionality between volume and absolute temperature, the 
new temperature (T') must be greater than the original 
temperature. The volume-fraction thus arranged to 
bring about this proportional increase when multiplied 
by the original temperature (T = 273° + 17° = 290°) 
becomes 250/100, and the final temperature accordingly 
is T' = 290 (250/100) or 725°. 725° A. is equal to 452° C. 


THE EFFECT OF TEMPERATURE UPON GASES 19 

The Relation of Density to Temperature. — In con¬ 
sideration of the change in absolute density under 
change of temperature, with pressure a constant, we see 
from the formula D = M/Y that, as the volume increases, 
the fraction M/Y (the expression for the density) must 
become less and less. With the volume of a gas — the 
denominator of this fractional quantity — becoming less, 
we obtain a larger quotient or a larger value for the den¬ 
sity. Now as this change in volume is directly propor¬ 
tional to a change in absolute temperature, the density of 
a gas will decrease with an increase in its absolute tem¬ 
perature and increase with a decrease in this factor, — 
that is, under constant pressure the absolute density of a 
definite quantity of gas is inversely proportional to its 
absolute temperature or Dj D / =J I' : T. 

Example 7. — A volume of hydrogen at 0° C., with the abso¬ 
lute density 0.08987, will have what density when expanded by 
heating to 10° C. at constant pressure? 

From the equation above we have D' = D (T/T'). 
This signifies that the density given must be multiplied 
by a fraction formed from the two temperatures (the 
temperature-fraction). We should now ask the question 
whether, from the inverse proportionality just mentioned, 
the new density (D') will be greater or less than the 
original density (D). Here the new temperature is 
higher, consequently the new density will be lower, there¬ 
fore the smaller number must be placed in the numerator 
and the value for D' can be readily calculated: 


D' = D (273/283), or 0.08987 (273/283), or 0.0867. 


20 


CHEMICAL CALCULATIONS 


PROBLEMS. 

13 . What will be the volume of 250 c.c. of hydrogen meas¬ 
ured at 30° when cooled to — 10° at constant pressure? 

Ans. 217 c.c. 

14 . A rubber balloon containing 400 c.c. of oxygen measured 

at — 20° is subjected to a temperature of 120°. What is the 
increase in volume of the balloon, the atmospheric pressure 
constant throughout? Ans. 221.3 c.c. 

15 . A volume of gas measuring 500 c.c. at 0° was expanded 
by heating to 600 c.c., at constant atmospheric pressure. What 
was the temperature which the gas attained? Ans. 54.6°. 

16 . The absolute density of oxygen is 1.429. When a given 
volume of this gas is warmed from 0° to 40° under a constant 
pressure, what will the absolute density of the gas become? 

Ans. 1.246. 

17 . A volume of gas with the absolute density 4 and measur¬ 

ing 250 c.c. at 0° was expanded by warming, under constant 
pressure, to a volume of 600 c.c. What increase in temperature 
was required, and what would be the weight of 300 c.c. of the 
rarefied gas? Ans. 382.2°; 0.5 gram. 


CHAPTER V. 


THE COMBINED EFFECT OF TEMPERATURE AND 
PRESSURE UPON GASES. PARTIAL PRESSURES. 

When a volume of gas is subjected to change in both 
temperature and pressure the relation of the original 
volume to the new volume must be in accordance with 
the two principles already enunciated: 

V' : V = P : P' and V' : V = T' : T 

or V'/V - P/P' and V'/V = T'/T 

or V' = V (P/P') and V' = V (T'/T). 

Considered separately, we know that the original volume 
under change in pressure must adjust itself in accordance 
with the ratio denoted by Boyle’s Law and attain that 
value which is expressed by the product of the pressure- 
fraction by the original volume, — V (P/P'). The 
operation of a change in temperature will modify the 
original volume in accordance with the Law of Charles, 
and the resulting value will be expressed, as the formula 
above indicates, by the product of the original volume 
by the temperature-fraction, — V (T'/T). 

If the original volume, already modified to accord with 
a change in pressure, is to come in turn under the influence 
of a change in temperature, which is separate and inde¬ 
pendent of the pressure, we shall be able to express the 
result by substituting this already modified value of the 
original volume, V (P/P'), for that volume denoted by 
V in the equation, V' = V (T'/T), representing the effect 
of an alteration in temperature. In other words, the vol- 


22 


CHEMICAL CALCULATIONS 


ume V in the equation V 7 = V (T'/T) may be regarded as 
already affected by some change in pressure, — a change 
which can have altered it only in accordance with the 
Law of Boyle and brought it to its new value through the 
expression V (P/P'). It follows, therefore, that by sub¬ 
stituting V (P/P') for V in the equation V' = V (T'/T) 
we shall obtain the value which results from the action of 
both these agencies, pressure and temperature, upon the 
gas. Whether we consider the action of pressure before 
or after that of temperature, the final product will be the 
same. Thus, when the expression V' = V (T'/T) is made 
to include the effect of a change in pressure upon the 
original volume V, we have the form 

V' = V(P/P') (T'/T),or V' = v(p . J); 

whereas the expression denoting the influence of pressure 
alone, V' = V (P/P'), when made to include the effect of 
a change in temperature upon the volume V, gives the 
same form 

V' = V (T'/T) (P/P'), or V' = V (|, . £) • 

Though the Law of Boyle holds only when temperature 
is a constant and the Law of Charles only when pressure 
is a constant, the simultaneous validity of the two laws is 
here assumed. This is more clearly understood when 
variations in temperature and pressure are studied under 
constant volume. Thus a volume of gas V, at tempera¬ 
ture T, and under a constant pressure P, when warmed to 
the temperature T', gives a new volume V', according to 
the equation V' = V (T'/T). In order to bring back this 
increased volume to its original volume V, under constant 
temperature, a pressure P' must be now applied. This 
pressure will be derivable from the original pressure in 
accordance with the Law of Boyle; and from the new vol- 


TEMPERATURE AND PRESSURE UPON GASES 23 


ume V', or Y (T'/T), we can only obtain the original or 
smaller volume V at the condition of this greater pressure 
P'; hence the equation P'V = V (T'/T) P, or that equa¬ 
tion which denotes the inverse proportionality between 
volume and pressure. With the volume V', thus brought 
back to volume V, through this increase in pressure, the 
variations between temperature and pressure at this con¬ 
stant volume can be derived. The elimination of the 
common factor V from the equation above gives us 


P' = P (T'/T), or P'/P = T'/T. 


This signifies that at constant volume the change in 
pressure is directly proportional to the change in tem¬ 
perature expressed upon the absolute scale, and is there¬ 
fore exactly analogous to the change in volume by 
change in temperature at constant pressure; the two 
thereby stand in harmony with the Law of Charles. 
The influence of temperature may be as well considered 
along with alterations in pressure as with volume, since 
the alterations are exactly analogous and are derived 
from identical expressions. Consequently the variations 
in the two factors, pressure and volume, may simulta¬ 
neously include the effect of temperature and take that 
form which indicates the direct proportionality of each 
of these factors to the absolute temperatures: 


T'/T = P'/P X V'/V. 

As more generally expressed, the equation is 
V'/Y = P/P' X T'/T, or 



This is, as seen, the product of the original volume by 
both the pressure-fraction and temperature-fraction con¬ 
jointly. Hence we may consider the original volume as 
having come under the action of two forces simultane- 


24 


CHEMICAL CALCULATIONS 


ously; the new volume, therefore, is that volume produced 
by the resultant of these two forces.* 

The Relation of Volume to Temperature and Pressure. 

Example 8. — Given 100 c.c. of a gas at 10° and 750 mm. 
pressure, what will be its volume at standard conditions (0° 
and 760 mm.)? 

The change from the original to the new values may be 
indicated thus: 283° —» 273°, and 750 mm. —» 760 mm. 
With respect to these alterations in temperature and pres¬ 
sure the questions may now be asked in regard to the 
magnitude of the new volume brought about by each of 
these influences. From the change in temperature the new 
volume must be smaller, as the gas in cooling undergoes 
a contraction (direct proportionality), hence 273/283 will 
represent the temperature-fraction. From the change in 
pressure the new volume must be smaller, since the new 
pressure is larger and the volume therefore will be dimin¬ 
ished (inverse proportionality), hence 750/760 will repre¬ 
sent the pressure-fraction. Bringing these two influences 
together, the product of the two by the original volume 
will give the combined effect of the two changes upon this ■ 
volume: 


V' = 100 (273/283) X (750/760), or V' = 95.2 c.c. 


* If in this connection the form of equation V' = V (1 + at) is used, 
we shall arrive at a similar expression for change in pressure at con¬ 
stant volume: P' = P (1 + at). The combined effect of a change in 
volume and change in pressure is then represented by P'V = PV 
(1 + at). By substituting here for t its value on the absolute scale, 
T — 273, we obtain 


'273 T - 273 
273 + 273 


j, or P'V 



P'V - PV 


The expression PV/273 is found to be a constant for a given quantity of 
gas under all values of temperature, pressure and volume, and is usu¬ 
ally expressed by R ; hence the general gas equation, PV = RT, which 
signifies “that the pressure-volume product of a definite quantity of 
any definite gas is proportional to its absolute temperature.” — 
A. A. Notes. 



TEMPERATURE AND PRESSURE UPON GASES 25 


With these two fractions each less than unity in value, 
the original volume is materially diminished under the 
new conditions. 

The Relation of Temperature to Volume and Pressure. 

Example 9 .—At what temperature will a gas 200 c.c. in 
volume, at 20° and 750 mm., occupy a volume of 300 c.c. 
when the pressure is 740 mm.? 

Here the volumes and pressures are known, and the 
undetermined quantity is one of temperature. The 
general formula must be arranged so that the undeter¬ 
mined quantity is on one side of the equation: 


V'/V = P/P' X T'/T 


when divided through by P / P' gives 

T'/T = V'/V X P'/P, 

hence the new temperature T' is expressed by the equation 



The entire matter may be simplified and made more 
apparent if we recall the reasoning advanced in the pre¬ 
ceding cases; namely, the unknown quantity is derivable 
from the known quantity by multiplying this latter by 
the fractions formed from the related terms, when these 
terms are arranged so as to give a quotient that will 
express the magnitude of the term sought (the unknown 
quantity) over that of the one already existent. 

At 20° C., or 293° A., the volume of gas in the problem 
above was 200 -c.c. At the new temperature it is to be 
300 c.c., hence this temperature must be larger, else no 
expansion could occur (direct proportionality); therefore 
T' = 293 (300/200). Again, the pressure-fraction must 
be arranged so that its quotient will acquire a value 
less than unity, i.e., 740/750, since the effect of a lower 


26 


CHEMICAL CALCULATIONS 


pressure must necessarily decrease the temperature here 
required (direct proportionality) (c/. p. 23). The com¬ 
bined influence will be expressed by the product of the 
two fractions by the original temperature: 


T' = 293 (300/200) X (740/750) 
T' = 433.6° A. = 160.6° C. 


or 


The Relation of Pressure to Volume and Temperature. 

Example 10 .— 100 c.c. of a gas at 20° and 740 mm. will 
exert what pressure when occupying 120 c.c. at 40° ? 

Arrange the related terms in fractions in accordance 
with the questions as to whether the new pressure is to be 
greater or less than the original, 740 mm., and multiply 
both by this value. The temperature-fraction becomes 
313/293, since at higher temperature the pressure must 
be greater (direct proportionality); the volume-fraction 
becomes 100/120, since at increased volume the pressure 
must be less (inverse proportionality), therefore by 
arranging all these factors we have: 


P' = 740 (100/120) X (313/293), or P' = 658.7 mm. 


This is in accord with the general equation which may be 
drawn up for changes in both volume and temperature , 
upon the pressure of a gas: 



When the volume is a constant the equation takes the 
form indicated on p. 23: 



The Relation of Density to Temperature and Pres¬ 
sure.— The variation in the absolute density of a gas 
through the simultaneous variations in temperature and 


PARTIAL PRESSURES 


27 


pressure follows readily from what has been stated in 
regard to variation in volume under these same influences. 
In the study of absolute density, D = M/V, the inverse 
proportionality between the absolute density and the 
volume of a gas permits, in the general equation denoting 
the influence of temperature and pressure changes in this 
connection, of the substitution of the term D 7 (the new 
density) and D (the original density) for Y and V' re¬ 
spectively. The general equation V'/V = P/P 7 X T 7 /T 
then becomes D/D 7 = P/P 7 X T 7 /T or, as transposed to 
indicate the alteration in temperature and pressure upon 
the original density: 



This is in keeping with what has been already observed, 
i.e.j the absolute density of a definite quantity of gas 
varies directly with the pressure and indirectly with the 
absolute temperature. 

Example 11. —A volume of oxygen at 20° and 750 mm. pres¬ 
sure has what absolute density under these conditions? 

The absolute density of oxygen at 0° and 760 mm. 
pressure is 1.429. The new density will be less under 
the conditions named, since both the pressure-fraction 
750/760, and temperature-fraction 273/293, will reduce 
the original value according to the equation: 

D 7 = 1.429 (273/293) (750/760), or D 7 = 1.314. 

PARTIAL PRESSURES. 

Whether we have to deal with a simple gas or a mixture 
of several gaseous components, the Law of Boyle is equally 
applicable. Each component of a gaseous mixture exerts 
a definite individual pressure, the same that it would exert 
were it alone present in this volume. According to the 
Law of Dalton, the sum of these individual pressures, 


28 


CHEMICAL CALCULATIONS 


'partial pressures, of the several components is equal to the 
total pressure of the mixture. When two gases of equal 
volume at the same conditions are brought together and 
communication made between the two vessels, we find 
that the gas from each vessel diffuses into the other vessel 
and becomes uniformly distributed throughout this double 
volume (the two vessels). In other words, each gas be¬ 
haves as if it alone were present in the total space included 
in the two vessels. Under double the original volume each 
gas can exert but one-half of its original pressure. The 
sum of the pressures of the two gases, however, must be 
again equal to unity or that pressure which each originally 
exerted. Consequently the total pressure of a gaseous 
mixture may be regarded as the sum of the individual or 
partial pressures of the several components. 

When aqueous vapor is present in a volume of gas we 
have again the consideration of gaseous mixtures. Water 
gives off a varying amount of its vapor, independent of 
any other gaseous substance that may be present in the 
space about it, but always dependent upon the tempera¬ 
ture. At any one temperature we assume an equilibrium 
between the two tendencies, — that of the molecules of 
the vapor to condense as liquid, and that of the molecules 
of the latter to fly off as vapor; in other words, we have a 
condition of saturation with aqueous vapor. 

The density or concentration of aqueous vapor (the aque¬ 
ous tension) as attained at each condition of equilibrium 
is definite for that temperature at which the equilibrium 
exists. This density of the vapor is determined most easily 
by measuring the pressure which it exerts, — the vapor 
pressure. If a little water is admitted into the vacuum 
at the top of a barometric column of mercury, the vapor 
evolved will exert a pressure which increases with a rise in 
the temperature of the surrounding medium. At 100° the 


PARTIAL PRESSURES 


29 


pressure of this vapor will have just sufficed to displace all 
of the mercury (sustained by the atmosphere at sea level), 
and consequently will have reached a pressure exactly bal¬ 
ancing one atmosphere. This temperature is the boiling- 
point of the liquid or that point at which bubbles of vapor 
form in the liquid itself. At higher elevations than sea 
level, i.e., under reduced atmospheric pressures, this 
attainment of the external pressure by the vapor of the 
liquid (the boiling-point) will take place at lower tempera¬ 
tures. As the temperature falls the aqueous vapor above 
the mercury in the barometric tube just mentioned will 
decrease in concentration and exert less pressure; conse¬ 
quently mercury will be forced into the tube by the pres¬ 
sure of atmosphere from without. The difference in height 
of the mercury in this tube and that in a barometer, where 
no moisture is present, will give the height of mercury, or 
pressure, corresponding to that of the aqueous vapor at 
the recorded temperature. 

When any gas is measured over water at atmospheric 
pressure, and a temperature constant for both gas and 
liquid, we obtain the partial pressure of the dry gas in the 
mixture by subtracting from this barometric reading the 
pressure which aqueous vapor exerts at the observed tem¬ 
perature. (The values for a range of temperatures are 
found in Appendix II.) 

Example 12. —100 c.c. of a gas at 10° and 750 mm., measured 
over water, will have what volume under standard conditions? 

At 10° the pressure of aqueous vapor is 9.2 mm. (see 
Appendix II), therefore the barometric reading, 750 mm., 
represents the sum of the pressures of the aqueous vapor 
and that of the dry gas, and, as standard conditions are 
desired, the volume of the gas should be determined in the 
dry state. Accordingly 750 mm. — 9.2 mm. = 740.8 mm., 
the pressure of the dry gas. The problem then resolves 


30 


CHEMICAL CALCULATIONS 


itself into one exactly as in Ex. 8, page 24. The volume 
of a gas at 740.8 mm., when calculated at a pressure of 
760 mm., or a greater pressure, must be necessarily 
diminished; therefore 740.8/760 is the pressure-fraction. 
The temperature-fraction is 273/283, consequently the 
new volume will be derived from the equation 

V' = 100 X 740.8/760 X 273/283, or V' = 94.03 c.c. 

PROBLEMS. 

18. 500 c.c. of hydrogen at 25° and 745 mm. pressure will have 
what volume at 15° and 755 mm. pressure? Ans. 476.8 c.c. 

19. A vessel of 2000 c.c. capacity held 5 grams of a vapor 
at the standard conditions of temperature and pressure. What 
weight of this vapor at 10° and 750 mm. pressure can be held in 
this vessel, the capacity considered constant? 

Ans. 4.76 grams. 

Note. —This volume of vapor (cc), at 0° and 760 mm., with the 
weight ( w ), when raised to the new volume ( y ), will still have the same 
weight. Consequently the vessel can he made to contain only that 
part of the original weight which is denoted by the fraction x/y. 

20. 10 liters of a gas measured at 20° and 750 mm. pressure 
weighed 14 grams. What weight of this gas could be contained 
in a smaller vessel holding 4 liters at 10° and 760 mm. pressure? 

Ans. 5.87 grams. 

21. What volume will 1000 c.c. of gas at 30° and 740 mm. 
pressure occupy when reduced to standard conditions? / 

Ans . 877.3 c.c. 

22. 1000 c.c. of a gas measured at 10° and 750 mm. pressure 
was increased to 1120 c.c. by warming. The final pressure read 
740 mm. What was the final temperature of the gas? 

Ans. 39.7°. 

23. A volume of gas measured at 10° and 750 mm. pres¬ 
sure will have what pressure at 20°, the volume a constant? 

Ans. 776.5 mm. 

24. A volume of gas measuring 1000 c.c. at 15° and 
745 mm. pressure was warmed to 32°. What increase in the 
pressure of the gas would be recorded, the volume a constant? 

Ans. 44 mm. 


TEMPERATURE AND PRESSURE UPON GASES 31 

25. What decrease in temperature will be necessary to reduce 

400 c.c. of a gas, at 20° and 765 mm., to a volume of 300 c.c. at 
750 mm.? Ans. 77.56°. 

26. What increase in temperature will be necessary to bring 
a volume of gas, measuring 560 c.c. at 10° and 745 mm. pres¬ 
sure, to a volume of 600 c.c. at this same pressure? 

Ans. 20.2°. 

27. A volume of hydrogen measuring 500 c.c. at 25° and 

730 mm. was reduced in volume to 400 c.c. at 0°. What was 
the final pressure of the gas? Ans. 836 mm. 

28. What increase in pressure is necessary to force 100 c.c. of 
hydrogen, at 10° and 740 mm., into a vessel of 80 c.c. capacity, 
when the temperature of this vessel is constant at 0°? 

Ans. 152.3 mm. 

29. What is the absolute density of hydrogen at 20° and 

740 mm. pressure? Ans. 0.08153. 

30. What is the absolute density of air at 10° and 750 mm. 

pressure? The absolute density at the standard conditions is 
1.293. Ans. 1.231. 

31. 400 c.c. of gas, with the absolute density 6 and measured 
at 25° and 750 mm. pressure, is to be brought to a temperature 
of 10° and a pressure of 760 mm. What weight of this final gas 
can be contained in a vessel of 100 c.c. capacity? 

Ans. 0.64 gram. 

32. 10 grams of a gas, measured at — 48° and 600 mm. pres¬ 
sure, was expanded by heating to 177° and reducing the pressure 
to 480 mm. Of this rarefied gas 250 c.c. weighed 0.5 gram. 
What was the original volume of gas and what was the density of 
the gas at its original and final volume? 

Ans. 2000 c.c. 

Original density = 5. 

Final density = 2. 

33. 150 c.c. of air measured over water at 18° and 746 mm. 
pressure will have what volume at standard conditions? 

Ans. 135.3 c.c. 

34. 1 liter of oxygen at standard conditions weighs 1.429 
grams. 440 c.c. of this gas measured over water at 24° and 
742 mm. pressure will contain what weight of the dry gas? 

Ans. 0.547 gram. 


32 


CHEMICAL CALCULATIONS 


35. 545 c.c. of nitrogen as measured over water at 22° and 

748 mm. pressure contain what weight of the dry gas (density = 
1.2507)? Ans. 0.6045 gram. 

36. 2.2 grams of oxygen will occupy what volume at 20° and 

770 mm. pressure when transferred to a vessel inverted over 
water? Ans. 1667 c.c. 

37. 1 gram of hydrogen (density = 0.08987) measured at 
standard conditions will occupy what volume when transferred 
to a vessel over water at 0° and 760 mm. pressure? 

Ans. 11,195 c.c. 

38. 400 c.c. of oxygen at standard conditions will have what 
volume when measured over water at 20° and 755 mm. pressure? 

Ans. 442.3 c.c. 

39. 100 c.c. of a gas, measured over water at 25° and 745 mm. 

pressure, will have what volume when deprived of moisture? 
The atmospheric conditions constant. Ans. 96.8 c.c. 

40. 1000 c.c. of oxygen, measured over water at 10° and 
750 mm. pressure, will have what volume at — 10° and 770 mm. 
pressure when deprived of its moisture? Ans. 894.1 c.c. 

41 . 500 c.c. of a gas contained in a tube inverted over 

water, and measured at 10° and 765 mm. pressure, will have 
what volume under a change in the atmospheric conditions to 
20° and 745 mm. pressure? Ans. 537.7 c.c. 

42. What decrease in pressure will be necessary to raise a 

volume of gas measuring over water 400 c.c., at 22.5° and 
748 mm. pressure, to a volume of 440 c.c. under the same 
conditions? Ans. 66.2 mm. 

Note. —The actual atmospheric pressure of the moist gas required 
for the new conditions will be obtained by adding the tension of aque¬ 
ous vapor to the value for P'. 

43. What increase in atmospheric pressure will be necessary 
to reduce 200 c.c. of a gas, measured in a tube over water at 10° 
and 720 mm., to a volume of 100 c.c. at 20° in this same tube? 

Ans. 769.2 mm. 

Note. —The calculated pressure for the dry gas must be increased 
by the tension of aqueous vapor at the new conditions in order to 
obtain the actual pressure that would be recorded (c/. Prob. 42). 


CHAPTER VI. 


AVOGADRO’S HYPOTHESIS AND SOME OF ITS 
APPLICATIONS. 

The hypothesis of Avogadro is the outcome of the Law 
of Combining Volumes (Gay-Lussac), and assumes that, if 
the Laws of Boyle and Charles are strictly true, there 
must be a uniform distribution of molecules in all volumes 
of gases at the same conditions of temperature and pres¬ 
sure. More generally the hypothesis takes the following 
form: Under the same conditions of temperature and 
pressure equal volumes of all substances in the state of 
vapor contain an equal number of molecules. By mole¬ 
cules are meant the smallest parts in which a substance 
maintains its identity. The complete chemical combina¬ 
tion between the molecules of equal or multiple volumes 
of gases to form definitely related volumes of gaseous 
products, with no molecules of either constituent remain¬ 
ing free or uncombined, presupposes this even distribution 
of molecules. 

Now the weight of a given volume of gas when compared 
with the weight of an equal volume of another gas, under 
the same conditions, will represent not alone the ratio 
between the sum of the weights of all the molecules in one 
volume and that of the other, but also the ratio between 
the weight of an individual molecule of one gas and that 
of the other. This follows, of course, from the fact that 
the numerical ratio between the weights of each equal 
volume is not altered through division by a common factor, 
— the unknown yet equal number of molecules in both 
33 


34 


CHEMICAL CALCULATIONS 


volumes. In determining the relative weights of mole¬ 
cules it is therefore of no importance as to what particular 
volume of gas is weighed, so long as the weight is to be 
compared with the weight of an exactly equal volume of 
some other gas at the same conditions of temperature and 
pressure. Any known volumes of gases upon which accu¬ 
rate data are at hand may serve for this purpose when 
they are reduced to the same conditions (c/. Chap. Y) and 
comparison made between equal volumes of each. 

The Relation of Density to Molecular Weight. — This 
comparison in the weights of equal volumes of gases is, 
after all, nothing more or less than the comparison in gas 
densities. Thus the absolute density of hydrogen, 0.08987, 
when compared with that of oxygen, 1.429, is found to be 
about 1/16 of the latter, or, if the density of hydrogen 
is considered as unity, the density of oxygen becomes 
15.90: 

0.08987 : 1.429 = 1 : 15.90. 

The relative weight of a ftiolecule of oxygen, — its molecu¬ 
lar weight, — would be, accordingly, 15.90 times the 
molecular weight of hydrogen. But from accurate deter¬ 
minations of the atomic weights of these elements the 
molecular weight of oxygen is found to be only 15.88 
times that of hydrogen.* 

* The determination of-the atomic weights of all the elements gives 
us the most accurate means of obtaining the true molecular weights. 
Density determinations involve numerous complications, and are there¬ 
fore liable to errors, but nevertheless they give us a very fair approach 
to the true values. The various degrees of cohesion between the 
molecules of gases bring into existence a greater or less deviation from 
the uniform packing of these molecules as assumed by Avogadro’s 
hypothesis for a perfect gas. Thus with oxygen we have a slightly 
greater packing than with hydrogen, as is shown by the increase of 
the relative density 15.90 over the calculated value 15.88. In order 
that our calculations may be free of these slight discrepancies, densi¬ 
ties may be given as calculated back from the correct molecular weights, 


AVOGADRO’S HYPOTHESIS 


35 


The relation in the absolute densities applies equally 
well to the relative densities of gases providing, of course, 
that they are referred to the same standard. Thus the 
relative densities of hydrogen and oxygen upon air as a 
standard are 0.0696 and 1.105 respectively. The com¬ 
parison of these values gives the same relative weights 
of the molecules of hydrogen and oxygen as shown 
above. 

We have several cases which show that a volume of 
hydrogen may enter into combination with an exactly 
equal volume of another gas, and the resulting gaseous 
compound occupy the volume originally held by the two 
gases severally, i.e., the sum of the two gaseous volumes 
now makes up the volume of the compound. Since 
temperature and pressure remain the same throughout, 
the number of molecules in the volume of the resulting 
compound must be twice the number of molecules in 
either volume of gaseous constituent. As each of the 
molecules in the compound contains hydrogen in chemical 
combination, we clearly see that there are now twice as 
many hydrogen parts present as there were in the original 
volume of hydrogen. Since weight , for weight there can 
be no change in the amount of hydrogen present in the 
two cases, we naturally infer that the original molecule of 
hydrogen contained two smaller parts or units, — atoms , — 
and that it was these atoms which were concerned in 
the chemical combination. Never have we been able to 
get the element hydrogen to spread out over more than 
twice its original volume when entering into chemical 
combination, therefore we assume that the hydrogen as 
we know it contains at least two atoms to its molecule. 

and consequently the value for oxygen will henceforth he taken as 
15.88 instead of 15.90, when referred to hydrogen as unity. 

Unless otherwise stated, the densities hereafter considered are the 
calculated weights of 1 liter of gas at standard conditions. 


36 


CHEMICAL CALCULATIONS 


We have a number of elements the molecules of which 
under certain conditions are shown in like manner to 
possess two atoms (i.e., diatomic), e.g ., oxygen (0 2 ), 
chlorine (Cl 2 ), bromine (Br 2 ), iodine (I 2 ), and nitrogen (N 2 ), 
whereas the molecule of mercury in the vapor state is 
monatomic (Hg). 

As already stated, from the comparison in the weights 
of equal volumes of gases, we may derive the relative 
weights of the several kinds of molecules. With the 
hypothesis now that the molecules themselves are com¬ 
posed of atoms, we must interpret these relative values as 
those which stand for the ratios between the sums of the 
weights of the atoms in the molecules, i.e., the molecular 
weights. A comparison between the relative weight of a 
molecule of hydrogen and a molecule of oxygen gives us 
the ratio 1 : 15.88. Since both hydrogen and oxygen 
molecules contain two atoms each, this same ratio stands 
likewise for that between the weights of the hydrogen and 
oxygen atoms. In order, therefore, to include the relative 
weights of the atoms in the molecular weights and thus be 
able to compare the relative weights of the molecules upon 
their smallest units as a basis, it is only natural that we 
regard the ratio 1 : 15.88 as the ratio between the relative 
weights of the hydrogen and oxygen atoms. The molecu¬ 
lar weights of hydrogen and oxygen then become 2 and 
31.76 respectively, values which indicate, of course, that 
they are made up of the sums of the respective atomic 
weights in each molecule. All of this presupposes that 
hydrogen, with the smallest atomic weight, should be re¬ 
garded as unity for the convenience of these comparisons. 

The importance of the element oxygen with its great 
number of well-defined compounds and the comparative 
ease with which these compounds may be utilized, through 
analyses, for the determination of the relative weights of 


AVOGADRO’S HYPOTHESIS 


37 


molecules of other elements, has led to its universal adop¬ 
tion as the standard of molecular weights. In order that 
the values upon this scale may not depart far from those 
already determined upon the basis of hydrogen as unity, 
the former value for oxygen, 31.76, was increased to 32, a 
whole number, and thus the atomic weights of all the 
elements allowed to retain a value greater than unity. 
With oxygen changed from 31.76 to 32, we may readily 
derive the corresponding value for the molecular weight 
of hydrogen from the simple proportion: 

31.76 :2 = 32 : x, 

which gives the value 2.016. From the molecular weight 
of oxygen (32) the relative weight of the atom must be¬ 
come 16, while the relative weight of the hydrogen atom 
will be 1.008. An imaginary gas with the atomic weight 
1 /16 of the atomic weight of oxygen (16) may be regarded, 
therefore, as the basis or unit weight of these values. 
These numerical values refer to no particular standard of 
weights, but when expressed in grams it is customary to 
call them gram-molecular weights; thus 32 grams is the 
G.M.W. of oxygen. 

As already stated, the comparison of the densities of all 
substances in the state of vapor gives at once the relative 
weights of their molecules. In order, then, to obtain the 
molecular weight of any substance upon the basis of oxy¬ 
gen as 32 it becomes necessary to effect a comparison 
between its absolute density and that of oxygen on this 
basis. The calculated absolute density of chlorine is 
3.166 (actual value = 3.22). The absolute density of 
oxygen, 1.429, will bear the same relation to this value 
for chlorine as the weight 32 bears to the molecular weight 
of chlorine, 70.92: 

1.429 : 3.166 = 32 : x, or x = 70.92. 


38 


CHEMICAL CALCULATIONS 


When the relative density of a gaseous substance is 
given, i.e., the value obtained by referring the absolute 
density to that of oxygen as unity, the proportion above 
will of course take the form: 

1 : Rel. density of gas = 32 : x. 

This proportion shows that the molecular weight ( x ) is 
merely the product of the relative density by 32. This 
is the logical outcome of making oxygen the standard 
both of gas densities and molecular weights. Since as the 
basis of relative densities the weight of a unit volume of 
oxygen is considered unity and as the basis of molecular 
weights it is considered 32, we naturally need only multiply 
any particular relative density by 32 in order to obtain 
the molecular weight of this substance. 

If, on the other hand, the relative density of a gas is 
given in terms of some other density besides oxygen as a 
standard, the molecular weight of this standard substance 
must of course replace 32 in the proportion above. For¬ 
merly the relative densities were determined upon the 
basis of hydrogen as unity. The proportion indicating 
the molecular weight upon this basis is as follows: 

1 : Rel. density (H = l) = 2.016 : x. 
Consequently the molecular weight may be derived from 
the relative density (H = l) by multiplying this value 
by 2.016. 

The Calculation of Relative Densities upon Different 
Standards. — To convert the relative density of a gas 
upon the hydrogen standard over to the standard of oxy¬ 
gen, or vice versa, we shall need to refer this value to the 
ratio between the absolute densities of the two standards, 
namely 0.08987 and 1.429, in an inverse order. As the 
means and extremes of the proportion we must have 
the product of the relative density of a substance by the 


AVOGADRO’s HYPOTHESIS 


39 


absolute density of the corresponding standard, a product 
always equal to the absolute density of the substance in 
question; and hence a constant for the proportion. Thus: 

I Abs. densityl (Abs. density] (Rel. density] [Rel. density] 
oxygen hydrogen |=| substance || substance | 
(1.429) J l (0.08987) J [ (H = l) ] ( (0 = 1). ) 

Example 13 .—The calculated relative density of chlorine 
(H = 1) is 35.23. What is the relative density of this gas 
referred to oxygen? 

According to the proportion: 1.429:0.08987 = 3^.23 : x, 
the value of x , or relative density, (O = 1), is calculated 
as 2.216. When this is multiplied by 32 we obtain the 
molecular weight of chlorine, 70.92. 

Quite often the vapor density of a substance is deter¬ 
mined upon air as a standard. The ratio 1.429 : 1.293 is 
the ratio between the absolute densities of oxygen and air 
respectively, hence the calculated molecular weight of air 
referred to oxygen as 32 will be derived from the propor¬ 
tion: 1.429 • 1.293 = 32 : x. From this the value 28.955 
is found; a value which signifies that if air were a com¬ 
pound its molecular weight would be 28.955. 

The molecular weight of a substance may be calculated, 
therefore, from its vapor density referred to air if we 
substitute this value, 28.955, for 32 in the proportion on 
page 38) and the relative vapor densities are made one 
member of the proportion. 

Example 14 -—The calculated vapor density of mercury is 
6.908 referred to air. What is its molecular weight? 

The proportion 1 : 6.908 = 28.955 : x gives 200 as the 
molecular weight of mercury. The conversion of any 
vapor density (air = 1) to the basis (O = 1) is exactly 
analogous to the case of hydrogen and oxygen above. 
The absolute density of air is 1.293. 


40 


CHEMICAL CALCULATIONS 


The Gram-Molecular Volume. — If the actual weight of 
1 c.c! of a gas is known in grams and this value is divided 
into the molecular weight also expressed in grams, the 
quotient will indicate the exact number of cubic centi¬ 
meters of the gas in question that will be required to give 
this weight or the gram-molecular weight (G.M.W.). In 
other words, the quotient expresses the volume in cubic 
centimeters which contains the gram-molecular weight. 

For example, 32 = G.M.W. of oxygen; 0.001429 gram 
is the weight of 1 c.c. of oxygen, therefore 

qo 

-= 22,390 (approximately), 

0.001429 ’ v ^ 

the number of cubic centimeters of oxygen, at 0° and 
760 mm., necessary to give its G.M.W. 

Again, the G.M.W. of hydrogen is 2.016 and the weight 
of 1 c.c. is 0.00008987 gram, therefore 


2.016 

0.00008987 


22,400 (approximately), 


the number of cubic centimeters of hydrogen at 0° and 
760 mm. necessary to give its G.M.W. 

This volume occupied by the G.M.W. of a substance 
is known as the gram-molecular volume (G.M.V.). Its 
calculation from the density of a gas gives values which 
diverge slightly from the average obtained for the more 
nearly perfect gases (i.e., .22,400 c.c.) according as the 
degree of packing of the molecules in each diverges from 
that in the latter. We deduce, therefore, the simple 
statement that the weight of 22,400 c.c. of a gaseous sub¬ 
stance, at 0° and 760 mm., will give, when expressed in 
grams, the gram-molecular weight of that substance. 
From the weight of any given volume of a gas the weight 
of 22,400 c.c. may be calculated by simple proportion; 




AVOGADRO’S HYPOTHESIS 41 

hence a ready means is given for determining molecular 
weights in general. 

Example 15 . —What is the molecular weight of a gas, 5600 c.c. 
of which at standard conditions weigh 5 grams? 

The value 22,400 is a standard, and the volume here to 
be compared with it (5600) must form the same ratio 
thereto as the relative weights of the two volumes: 

5600 : 22,400 = 5 : x, or = •>. 

22,400 x 

From which the molecular weight of this gas ( x ) is found 
to be 20. 

With this definite relation established between the 
volume 22,400 c.c., calculated at 0° and 760 mm., and that 
weight in grams which stands for the molecular weight 
of an element or compound, we may derive the fractional 
volume of this G.M.V. which any corresponding fractional 
part of the G.M.W. of a substance may occupy, and, vice 
versa, any fractional weight of the G.M.W. which any 
fractional part of this G.M.V. of the substance in state of 
vapor may have. In other words, the same fractional 
parts of the G.M.V. and the G.M.W. are directly pro¬ 
portional to each other. 

Example 16 . — 16 grams of oxygen will occupy what vol¬ 
ume at 0° and 760 mm. pressure? 

Since 32 grams occupy 22,400 c.c. at these conditions, 
the volume occupied by 16 grams will be 16/32 of 22,400 
or 11,200 c.c.,or, by simple proportion, 32 :16 = 22,400 : x , 
or x = 11,200 c.c. 

To apply the effect of temperature and pressure changes 
let it be desired to find the volume which 2 grams of 
nitrogen will occupy at 20° and 750 mm. The molecular 
weight of nitrogen is 28. 22,400 : x — 28:2, or x = 1600 c.c. 
This volume corrected for temperature and pressure be- 



42 


CHEMICAL CALCULATIONS 


comes increased at 20° by 293/273 of itself, and, at 750 mm., 
by 760/750 of itself, and hence the corrected volume will 

be 1600 

Vice versa, the weight of any volume of a substance in 
the state of vapor may be determined if the gram-molecu¬ 
lar weight of the compound is known. 

Example 17 .— What is the weight of 5600 c.c. of hydrogen 
chloride measured at 0° and 760 mm.? 


/293 760\ 

\273 750/ 


or 1739.2 c.c. 


The molecular weight of hydrogen chloride is 35.46 4* 
1.01, or 36.47. Therefore 

22,400 : 5600 = 36.47 : x, or x = 9.118 grams. 


Calculation of Densities of Gases from the Molecular 
Weights. — From these relations the determination of the 
density of any volume of vapor is reduced to the simple 
task of ascertaining from the molecular weight of the 
substance and this volume 22,400 c.c., what the weight 
of a unit volume (1 liter for gases) will be at standard con¬ 
ditions. The determination of the relative density of one 
gaseous substance upon any other as a standard resolves 
itself, therefore, into a comparison of the gram-molecular 
weights of the two gases, i.e., the comparison of the 
weights of these equal volumes, -— the G.M.V. With air, 
the weight of 22,400 c.c. has been given as 28.955 grams 
(this may be construed as the hypothetical gram-molecular 
weight of air). 

Example 18. — What is the relative density of chlorine re¬ 
ferred to (a) air? to (6) oxygen? 

(a) The gram-molecular weight of chlorine is 70.92, the 
weight of 22,400 c.c. The weight of 22,400 c.c. of air is 
28.955 grams. Therefore the ratio 70.92/28.955 when made 
equal to the ratio x/1 (where the weight of an equal volume 


AVOGADKO’S HYPOTHESIS 43 

of air is unity) gives us the proportion: 70.92: 28.955 = z : 1, 
or the relative density of chlorine as 2.449 (air =1). 

(b) When referred to oxygen the relative density of 
chlorine is determined in an exactly similar manner. The 
G.M.W. of oxygen, 32, replaces the weight of 22,400 c.c. 
of air in the example above. The process is of course the 
reverse of that in which the molecular weight of a sub¬ 
stance is derived by multiplying its relative density (0 = 1) 
by the value 32, the basis of molecular weights (0 = 32) 
(c/. p. 38). 


PROBLEMS. 

44. The relative density of carbon dioxide (H= 1) is 21.83. 

What is the relative density of this gas upon the oxygen stand¬ 
ard? Ans. 1.373. 

45. The relative density of chlorine (air = 1) is 2.449. What 
is its relative density upon the oxygen standard? 

Ans. 2.216. 

46. The vapor density of phosphorus trichloride (air = l) 
is 4.745. What is its molecular weight? Ans. 137.38. 

Note. — Though Ex. 14 may be followed it is much more simple 
to calculate the weight of 1 liter of the vapor, which is 4.745 times the 
weight of 1 liter of air (1.293 grams), and with this weight of 1 liter 
to apply Ex. 15 t 

47. What is the molecular weight of mercuric chloride, the 
vapor density of which is 9.354 referred to air? 

Ans. 270.84. 

48. The vapor density of water is 0.622 (air = l). What 

is its molecular weight? Ans. 18.016. 

49. What is the molecular weight of sulphur dioxide, the 
relative density of which is 2.0Q2 (0 = 1)? Ans. 64.07. 

50. What is the molecular weight of a gas, 2 liters of which at 

standard conditions weigh 12 grams? Ans. 134.4. 

51. The weight of 3840 c.c. of a certain vapor, at standard 

conditions, is 24 grams. What is the molecular weight of the 
substance? Ans. 140. 


44 


CHEMICAL CALCULATIONS 


52. 3180 c.c. of a gas measured at 24° and 750.2 mm. pressure 
weighed 6 grams. What is the molecular weight? Ans. 48. 

53. 8019 c.c. of a gas measured over water at 20° and 

742.4 mm. pressure weighed 14 grams when deprived of aque¬ 
ous vapor. Calculate the molecular weight. Ans. 44. 

54. 5647 c.c. of a gas measured over water at 24° and 
754.2 mm. pressure weighed 6.254 grams when deprived of aque¬ 
ous vapor. Calculate its molecular weight. Ans. 28.02. 

55. What volume will 49.63 grams of chlorine (mol. wt. = 
70.92) occupy at standard conditions? Ans. 15,680 c.c. 

56. What is the volume occupied by 8.8 grams of carbon 
dioxide (mol. wt. = 44) at 12° and 752 mm. pressure? 

Ans. 4726 c.c. 

57. 10 grams of carbon dioxide (mol. wt. = 44) will occupy 

what volume when contained in a vessel over water at 20° and 
742.4 mm. pressure? Ans. 5728 c.c. 

58. 1 gram of oxygen will occupy what volume over water 

at 30° and 756.5 mm. pressure? Ans. 814.3 c.c. 

59. 8 grams of hydrogen (mol. wt. = 2.016) are to be ad¬ 
mitted into a balloon at a temperature of 20° and a pressure of 
740 mm. What must be the capacity of the balloon? 

Ans. 97,980 c.c. 

60. Of what capacity is that vessel which contains 4 grams 

of oxygen (mol. wt. = 32) at the temperature of 18° and pres¬ 
sure of 748.4 mm.? Ans. 3031 c.c. 

61. 10.08 grams of hydrogen (mol. wt. = 2.016) at 0° and 

760 mm. pressure are to be forced into a vessel of 11.2 liters 
capacity. Under what pressure will the hydrogen be at this 
same temperature? Ans. 10 atmospheres. 

62. At what temperature will 8 grams of oxygen (mol. wt. = 

32) under a pressure of 760 mm. occupy a volume of 11.2 liters at 
this same pressure? Ans. 273°. 

63. A vessel holding 30.8 grams of carbon dioxide (mol. 
wt. = 44) at 10° and 740 mm. pressure is to be brought to a 
temperature of 50° and a pressure of 750 mm. What weight 
and volume of carbon dioxide will be lost? 

Ans. 2105 c.c. gas at latter conditions; 3.449 grams. 


AVOGADRO’S HYPOTHESIS 


45 


64. What is the weight of 6.594 liters of oxygen (mol. wt. = 

32) at 40° and 740 mm. pressure? Ans. 8 grams. 

65. What is the absolute density of hydrogen sulphide (mol. 

wt. = 34.09). Ans. 1.52. 

66. What is the relative density of hydrogen sulphide (mol. 
wt. = 34.09) referred to air and also to oxygen? 

Ans. 1.177 (air = 1). 

1.065 (O = 1). 

67. What is the relative density of hydrogen chloride (mol. 
wt. = 36.47) referred to hydrogen, air, oxygen and chlorine? 

Ans. 18.05 (H = 1). 

1.259 (air = 1). 

1.139 (O = 1). 

0.514 (Cl = 1). 

68. 8 grams of oxygen were mixed with 10.08 grams of 

hydrogen (mol. wt. = 2.016). Both gases were measured at the 
standard conditions. What was the relative density of this 
mixture (0 = 1)? Ans. 0.1076. 

69. A gas with the relative density 1.5 (0 = 1) was reduced 

from standard conditions to 20° and 740 mm. pressure so that 
it might have a volume measuring 6172.3 c.c. What is the 
weight of this volume of the gas? Ans. 12 grams. 

70. A volume of gas, with the relative density 0.8757 (0 = 1), 

was found to measure 1560 c.c. when transferred to a vessel 
over water at 18° and 742.4 mm. What is the weight of the 
dry gas here concerned? Ans. 1.75 grams. 


CHAPTER VII. 


THE LAW OF DEFINITE PROPORTIONS. 

When two or more elements unite in chemical combi¬ 
nation, a definite ratio is found to exist between the quan¬ 
tities of each element present and also between each of 
these several quantities and that of the molecular weight 
of the compound itself. In other words, upon the basis 
of 100 as the molecular weight, the amount by weight, 
here the percentage, of each element in the compound is 
constant and must bear a constant and unalterable ratio 
to the percentage by weight of any other element present. 

The molecular weight of a compound represents the 
sum of the atomic weights that go to form it. The atomic 
or unit weight, i.e., the symbol weight, is regarded as the 
smallest weight in which an element can be present in a 
molecular weight of any of its volatile compounds. A 
formula, therefore, is merely the expression in symbols 
which, under the existing conditions, denotes the number 
of atomic weights of the several elements in one molecu¬ 
lar weight; this is better called the molecular formula (c/. 
Chap. VIII). The multiples of these atomic weights 
which occur in the molecular weight of a compound are 
always expressed by integers which are usually small. 

The Percentage Composition of a Compound. — The 
percentage composition of a compound may be deter¬ 
mined by analytical or synthetical means. Thus, for 
example, 10 grams of magnesium unite with 6.58 grams 
of oxygen (when heated in an atmosphere of oxygen), 
and the resulting magnesium oxide weighs 16.58 grams. 

46 


THE LAW OF DEFINITE PROPORTIONS 


47 


Of the 16.58 grams magnesium oxide, 10 grams are of 
course magnesium, or the fractional quantity 10/16.58 rep¬ 
resents the weight of magnesium in magnesium oxide; 
the fraction 6.58/16.58 will represent, then, the propor¬ 
tional amount of oxygen present in the compound. These 
same definite ratios hold for all quantities of magnesium 
oxide; hence, if 100 grams are taken as a basis, 10/16.58 
of 100, or 60.3, will give the amount of magnesium in 
the 100 grams, i.e., the percentage of this element in the 
compound. Similarly, 6.58/16.58 of 100, or 39.7, is the 
percentage of oxygen present. By reference to simple 
proportion upon the basis of 100 as the total weight, 
these same percentage amounts are readily determined: 
MgO : Mg = 100 : x 
16.58 : 10 = 100 : x, or 60.3 per cent 

MgO : O = 100 : x 
16.58 : 6.58 = 100 : x, or 39.7 per cent. 

Now the formula of magnesium oxide is MgO, in which 
one atomic weight of magnesium (24.32) is combined with 
one atomic weight of oxygen (16). From the Law of 
Definite Proportions we know the ratios Mg/MgO and 
O/MgO are constant and, when expressed in numerical 
values of the several atomic weights concerned, become 
respectively: 

Mg or 24.32 or 2L32 
and MgO 24.32 + 16 40.32 

^ JL or 16 or 16 

MgO 24.32 + 16 40.32 ' 

The constant ratio 24.32/40.32 stands for the amount of 
magnesium in magnesium oxide in all amounts of the 
latter, and upon a percentage basis will be represented by 
this same fractional part of 100, i.e., 24.32/40.32 of 100, 
or 60.3 per cent. Similarly, for the amount of oxygen in 








48 


CHEMICAL CALCULATIONS 


magnesium oxide when expressed on a percentage basis, 
16/40.32 of 100, or 39.7 per cent, will be obtained. These 
values are identical (c/. preceding paragraph) with those 
obtained in the synthesis of this compound where a defi¬ 
nite weight of magnesium was converted into the oxide. 
In accordance with the Law of Definite /Proportions, the 
same relations must hold for all determinations in this 
compound. 

The general expression for problems of this type when 
the relative molecular weights are referred through simple 
proportion to a basis of 100 is as follows: 

MgO : Mg = 100 : x 
40.32 : 24.32 = 100 : (60.3) 

and 

MgO : O = 100 : x 

40.32 : 16 = 100 : (39.7). 

In the simple compound water, with the formula H 2 0, 
the quantity of hydrogen in one molecular weight is rep¬ 
resented by two atomic weights of this element. This 
aggregate of atomic weights, which represents the quan¬ 
tity in which any element, or group of elements, is present 
in a molecular weight of some compound, may be regarded 
as the formula-quantity for the particular element, or ele¬ 
ments, concerned in the total molecular weight. Thus 
2 H stands for the formula-quantity of hydrogen (2) in 
water (18), and the ratio 2 H/H 3 0 or 2/18 or 1/9 represents 
this constant proportion between the formula-quantity of 
hydrogen and the molecular weight of water when con¬ 
cerned, of course, in this compound.* As there is but one 

* The molecular weight of a substance is, after all, a formula- 
quantity for that particular group of elements when associated in this 
compound. The ratio between the two is always expressed by unity. 
Thus 18 is the formula-quantity of water in its molecular weight as 
steam, whereas in the form of ice the formula-quantity, just as the 
molecular weight, must be increased to 3 (?) times this number if the 
molecule is to be regarded as (H 2 0) 3 . 


THE LAW OF DEFINITE PROPORTIONS 49, 

atomic weight of oxygen in the molecular weight of water, 
the second ratio is simply 0/H 2 0 or 16/18 or 8/9. 
Hence in water 1 part by weight of hydrogen (1/9) is 
combined with 8 parts by weight of oxygen (8/9). Ex¬ 
pressed in percentage composition we have: 

H 2 0 : 2 H = 100 : * 

18.02 : 2.02 = 100 : 11.2 per cent hydrogen 

and 

H 2 0 : O = 100 : 88.8 per cent oxygen, 

the same ratio, 1 : 8, holding true, of course, for the 
percentages. 

In the compound sodium sulphate, Na 2 S0 4 , we may 
express the constant ratios: 2Na/Na 2 S0 4 , S/Na 2 S0 4 , and 
40/Na 2 S0 4 , by percentages of the molecular weight, 
142.07, or (46.0 + 32.07 + 64), as follows: 

Na 2 S0 4 : 2 Na = 100 : x 

142.07 : 46.0 = 100 : 32.38 per cent 

Na 2 S0 4 : S = 100 : x 

142.07 : 32.07 = 100 : 22.57 per cent 

Na 2 S0 4 : 4 O = 100 : x 

142.07 : 64 = 100 : 45.05 per cent 

The sum of these = 100 per cent. 

The Relation of Formula-Quantity to Molecular Weight. 

— Whatever the weight of a substance under considera¬ 
tion may be, the ratio between any one formula-quantity 
and the total molecular weight of the compound is always 
equal to the ratio between the actual weights which here 
correspond to these respective quantities. 

Example 19. — Calculate the weight of chlorine present in 
50 grams of sodium chloride. 


50 


CHEMICAL CALCULATIONS 


The formula of sodium chloride, NaCI, indicates that 
1 atomic weight of sodium (23.00) is combined with 
1 atomic weight of chlorine (35.46) to give a molecular 
weight of 58.46. The ratio Cl/NaCl or 35.46/58.46 
expresses the relation between the formula-quantity of 
chlorine and the molecular weight of the salt, and all 
weights of these substances when jointly involved in the 
compound salt must be in accord with this ratio. When 
this ratio is converted to a percentage basis we obtain the 
value 60.6 as the percentage of chlorine in all amounts of 
salt: 58.46 : 35.46 = 100 : 60.6. If, then, 60.6 per cent of 
any weight of salt is chlorine, 50 grams X 60.6 per cent or 
30.3 grams is the amount of chlorine present in the 
known weight of salt. 

Without resort to the percentage composition of a com¬ 
pound the constant ratio between the formula-quantity 
of any element present and the total molecular weight 
may be used directly for the determination of the amount 
of this element in any weight of compound. In the 
example just given, 35.46/58.46 represents the propor¬ 
tional amount of chlorine in salt. Without reference, 
then, to 100 as a basis, this same proportional part of the 
weight of salt taken must give the required value foi 
chlorine; thus 50 grams X 35.46/58.46 = 30.3 grams or 
weight of chlorine present; or, as is more customary, we 
may express the same by simple proportion: Since 58.46 
grams of salt contain 35.46 grams of chlorine, 50 grams of 
salt can contain only that proportional part of its own 
weight which 35.46 is of 58.46; the ratio 58.46 : 35.46 at 
once becomes equal to the ratio 50 : x; thus 

58.46 : 35.46 = 50 : x, or x = 30.3 grams. 

These proportions are most easily formed when we 
bear in mind that only one set of related terms, e.g., the 


THE LAW OF DEFINITE FEOPORTIONS 


51 


formula-quantities, constitute one member of the propor¬ 
tion, whereas the other set of related terms, the actual 
weights (here with one unknown factor), constitute the 
other member. 

Example 20. — Calculate the weight of sodium chloride 
that can be prepared from 30.3 grams of chlorine. 

The amount of salt that can be prepared from this 
definite weight of chlorine (30.3 grams) will be represented 
by the same ratio, Cl/NaCl, as under the preceding 
example (Ex. 19); but, for the purposes of multiplication 
into known values, it is necessary to use the inverted 
form NaCl/Cl or 58.46/35.46, with the large value in the 
numerator so that a proportional increase in the values 
affected by this ratio may be obtained. The same end 
may be reached by the use of simple proportion, where 
the unknown term increases over the known in direct 
ratio to the corresponding formula-quantities: 

35.46 : 58.46 = 30.3 : x, or x *= 50 grams. 

The Relation of Formula-Quantities to Each Other. — 

The ratio between any two formula-quantities occurring 
in one molecular weight may be used, of course, inde¬ 
pendently of the molecular weight, and when the ratio 
between certain groups of elements in a formula is to be 
determined the same principle will be found to apply 
as between single elements. The symbols representing 
these groups serve as usual for the calculation of the 
correct formula-quantities. In analyses of minerals the 
determination of these formula-quantities for various 
groups of oxides, etc., aids materially in the general classi¬ 
fication; thus in the mineral feldspar, AIK (Si 3 0 8 ), or 
Al 2 K 2 (Si 3 0 8 ) 2 , when intended for resolution into the con¬ 
stituent groups, becomes A1 2 0 3 . K 2 0.6 Si0 2 , from which 


52 


CHEMICAL CALCULATIONS 


the percentage of any one of the oxides present may be 
calculated from the corresponding formula-quantity. 

Example 21. — What weight of aluminium is present in a 
sample of pure cryolite, A1 2 F 6 . 6 NaF, which analyzed for 10 
grams of sodium? 

The ratio between the formula-quantity of each of the 
two substances concerned is 2 Al/6 Na, or 2 (27.1)/6 (23), 
or 54.2/138. As this is a constant we need only refer the 
ratio directly to the equivalent ratio when based upon 
sodium as 10 grams, — 54.2/138 = x /10, — or, as has been 
the practice, 138 : 54.2 = 10 : x, from which we ob¬ 
tain the value 3.93 grams for x, the weight of aluminium 
present. 

The Relation of Formula-Quantities to Percentage Com¬ 
position. — When data are given in percentages and it 
is desired to ascertain the percentage amount of some other 
element or group of elements present in the same com¬ 
pound, the calculations are made directly upon these per¬ 
centages. They occupy, of course, the identical relation 
to each other as do the actual weights themselves, but 
with the condition that a constant value of 100 serves as a 
basis for the entire molecular weight. 

Example 22. — What percentage of potassium oxide, K 2 0, is 
present in a sample of feldspar, AlK(Si 3 0 8 ), which analyzed 
for 12.2 per cent potassium? 

The ratio between the potassium and its corresponding 
amount of potassium oxide possible of existence in this 
same molecule is expressed by 2 K/K 2 0. With this defi¬ 
nite quantity of oxygen in the molecule now to be asso¬ 
ciated with the potassium, there naturally will be an 
increase in the percentage amount of the group K 2 0 over 
that of the potassium alone. The ratio between these 
percentage amounts must be in accordance with the corre- 


THE LAW OF DEFINITE PROPORTIONS 


53 


sponding formula-quantities, 2 K and K 2 0, upon which 
their weight and consequent percentage is dependent; 
therefore the ratio 2 K/K 2 0 must be equal to the ratio 
12.2% /x%. From the equation 2 K/K 2 0 or 78.2/94.2 = 
12.2/x we obtain, as the value of x, 14.7 per cent, i.e., 
14.7 per cent of a compound of the composition noted, 
and analyzing for 12.2 per cent potassium, may be con¬ 
sidered as potassium oxide. 

Percentage Composition as a Basis for Percentage 

Purity. — Upon this method of comparison it is a very 
simple task to calculate the percentage purity or per¬ 
centage amount of some compound in one of its samples 
submitted to an analysis; and this too upon the data 
secured in the determination of only one of its constituent 
elements. 

Example S3. — A sample of salt analyzed for 50 per cent 
chlorine. Assuming all of the chlorine to have been present as 
sodium chloride, what was the percentage of salt in the sample? 

The ratio Cl/NaCl must be equal to the ratio between the 
corresponding percentage amounts of these quantities,— 
50%/x%. Cl/NaCl = 35.46 : 58.46 = 50/a; or, by simple 
proportion, 35.46 : 58.46 = 50 : x. From this, x is found 
to be 82.4 in place of the theoretical 100. This is, there¬ 
fore, the percentage purity of the sample. When analyti¬ 
cal data are not given in percentages, the weight of both 
sample and final product must be known in order to deter¬ 
mine the percentage amount, of any constituent present. 

The percentage purity of a compound may be deter¬ 
mined also by calculating the weight of substance theo¬ 
retically possible from the results at hand (c/. Ex. 20) 
and then determining what percentage the weight of 
sample bears to this calculated value. 

The Interrelationship of two Formula-Quantities through 
a Common Quantity. — When the results of an analysis 


54 , CHEMICAL CALCULATIONS 

are given in terms of some compound other than that 
involved in the investigation it is necessary to calculate 
these results through the element common to both com¬ 
pounds. 

Example 24 . — A sample of feldspar, KAlSi 3 0 8 , weighing 
0.2507 gram gave upon analysis 0.0658 gram of potassium 
sulphate, K 2 S0 4 . What was the percentage of potassium oxide 
present in the sample? 

We have here two distinct ratios: K 2 S0 4 /2 K, which 
answers for the definite quantity of potassium present in 
this weight of potassium sulphate, and also 2 K/K 2 0 
for the quantity of oxide possible from a definite weight 
of potassium (cf. Ex. 22). Since the ratios involve the 
same definite quantities of potassium throughout, they 
may be considered as equivalent in respect to this ele¬ 
ment, and when brought together may be cleared of this 
common term, 2 K, as indicated when K 2 S0 4 /2 K is 
involved with 2 K/K 2 0. By multiplication we should 
have K 2 S0 4 /K 2 0, or that ratio with which the corre¬ 
sponding ratio between the actual weights of these 
substances, involving the same weight of potassium, 
must accord. From the known weight of potassium sul¬ 
phate, the actual weight of potassium oxide is easily 
calculated: K 2 S0 4 /K 2 0 or 174.27/94.27 = 0.0658/a:, i.e ., 
174.27 : 94.27 = 0.0658 : x , or x, the weight of potassium 
oxide in the sample, is equal to 0.0356 gram. From this 
weight of oxide and the original weight of the sample the 
percentage amount of potassium oxide in the compound 
may be calculated: 0.2507 : 0.0356 = 100 : x, otx= 14.2 
per cent. Without the simplification noted it would 
have been necessary first to determine the actual weight 
of potassium present in the potassium sulphate, and 
second, from the weight of potassium, the consequent 
weight of oxide. 


THE LAW OF DEFINITE PROPORTIONS 55 

This cancellation or elimination of a common term 
between two ratios may be illustrated by a second example 
somewhat more complicated. 

Example 25. — What percentage of potassium chloride can 
be regarded as present in a sample of potassium chlorplatin¬ 
ate, K 2 PtCl 6 , which analyzed for 43.6 per cent chlorine? 

The ratio of the formula-quantity of chlorine to that of 
potassium is 6 Cl/2 K. Now the ratio between potassium 
and potassium chloride in one molecular weight of potas¬ 
sium chloride is K/KC1. By examination of the formula 
of potassium chlorplatinate it is seen that two molecules 
of potassium chloride are involved in each molecule of 
chlorplatinate. This latter ratio then becomes 2 K/2 KC1 
when considered in the molecule of chlorplatinate. By 
this manner the same formula-quantity of potassium 
is considered in both ratios (6 Cl/2 K and 2 K/2 KC1); 
consequently this equivalent quantity may be eliminated 
from each (cf. Ex. 24), and the final ratio 6 Cl/2 KC1 ob¬ 
tained for the corresponding weights of chlorine and potas¬ 
sium chloride possible in a molecule of chlorplatinate. 
Though a fractional quantity of the total chlorine is 
brought into consideration as potassium chloride, this in 
no way disturbs the definite relation between the total 
chlorine and the total potassium even when considered 
as chloride. The ratio may be simplified, if desired, 
to 3 C1/KC1, and is equal to the ratio between the cor¬ 
responding percentage amounts of these substances, 
43.6 %/x%. From the equation 

3 C1/KC1 = 106.38 : 74.56 = 43.6/rr 

we obtain 30.56 as the percentage amount of potassium 
chloride possible in the sample. 


56 


CHEMICAL CALCULATIONS 


PROBLEMS. 

71. Determine the percentage composition of calcium car¬ 
bonate, CaC0 3 . Ans. Ca, 40.05%; C, 11.90%; O, 47.96%. 

72. Determine the percentage composition of ammonium 
nitrite, NaN0 2 . Ans. Na, 33.33%; N, 20.30%; O, 46.37%. 

73. Determine the percentage composition of potassium 
chlorate, KC10 3 . Ans. K, 31.90%; Cl, 28.93%; O, 39.17%. 

74. Determine the percentage composition of crystallized 
sodium *ulphate, NajSCL . 10 H 2 0. 

Ans. Na, 14.28%; S, 9.95%; O, 69.51%; H, 6.26%. 

75. Determine the percentage composition of sulphuric acid, 

H 2 S0 4 . Ans. H, 2.05%; S, 32.70%; O, 65.25%. 

76. Determine the percentage of sulphur trioxide, S0 3 , in 

sulphuric acid. Ans. 81.64 per cent. 

77. Determine the percentage of water in crystallized sodium 

carbonate, Na 2 C0 3 .10 H 2 0. Ans. 62.95 per cent. 

78. Determine the percentage composition of ordinary alum, 
K 2 A1 2 (S0 4 ) 4 .24 H JO. 

Ans. K, 8.24%; Al, 5.71%; S, 13.52%; O, 67.43%; H, 5.10%. 

79. What weight of sodium is present in 50 grams of sodium 

nitrate, NaNO s ? Ans. 13.53 grams. 

80. What weight of sodium is present in 100 grams of sodium 

hydrogen carbonate, NaHCO s ? Ans. 27.38 grams. 

81. What weight of oxygen is present in 200 grams of potas¬ 
sium chlorate, KC10 3 ? Ans. 78.34 grams. 

82. What weight of oxygen is present in 200 grams of mer¬ 
curic oxide, HgO? Ans. 14.815 grams. 

83. What weight of mercuric oxide, HgO, will contain 30 

grams of oxygen? Ans. 405 grams. 

84. What weight of potassium chlorate will contain 30 grams 

of oxygen? Ans. 76.6 grams. 

85. What weight of sulphuric acid can be prepared from 

100 grams of sulphur? Ans. 327 grams. 

86. Calculate the weight of potassium in a sample of pure 
sylvite, KC1, which analyzed for 2.230 grams of chlorine. 

Ans. 2.459 grams. 


THE LAW OF DEFINITE PROPORTIONS 


57 


87. What weight of copper is present in a sample of pure 

copper sulphate, CuS0 4 .5 H 2 0, which analyzed for 30.2 grams 
of sulphur trioxide, S0 3 ? Ans. 23.97 grams. 

88. Calculate the percentage purity of a sample of horn- 
silver, AgCl, which analyzed for 74.2 per cent silver. 

Ans. 98.61 per cent. 

89. Calculate the percentage purity of a sample of marble, 
CaC0 3 , which analyzed for 39.6 per cent calcium. 

Ans. 98.88 per cent. 

90. Calculate the percentage of potassium chloride in a sample 

of carnallite, KC1. MgCl 2 .6 H 2 0, which analyzed for 37.72 per 
cent chlorine. Ans. 26.44 per cent. 

Note. —Theory calls for 26.83 per cent KC1. Therefore the devia¬ 
tion, 0.39, between the percentage values means an error of 1.5 per cent 
from the theory (26.83). Halogen determinations are usually quite 
accurate. If the analysis is at all good, the salt falls a little short of 
purity. 

91. Calculate the percentage of calcium oxide, CaO, present 

in a sample of marble, CaC0 3 , which analyzed for 43.8 per cent 
carbon dioxide. Ans. 55.84 per cent. 

Sample practically pure; theory, 56.04 per cent. 

92. A sample of sodium chromate, Na 2 Cr0 4 , weighing 1.6780 

grams, gave upon analysis 1.4620 grams of sodium sulphate, 
Na 2 S0 4 . What was the percentage of sodium oxide, Na 2 0, in 
the sample? Ans. 38.01 per cent. 

Theory, 38.25 per cent. 

93. Calculate the percentage purity of a quantity of potas¬ 
sium ferrocyanide, K 4 Fe(CN) 6 , 0.5793 gram of which gave upon 
analysis 0.4650 gram of potassium sulphate, K 2 S0 4 . 

Ans. 84.8 per cent. 

Note. — From this weight of K 2 S0 4 determine the actual weight, 
and finally the percentage, of potassium in the sample and then apply 
Ex. 23. 

94. What is the percentage of potassium sulphate, K 2 S0 4 , in 
a sample of common alum, K 2 S0 4 . A1 2 (S0 4 ) 3 .24 H 2 0, which 
analyzed for 33.51 per cent sulphur trioxide, S0 3 ? 

Ans. 18.23 per cent K 3 S0 4 . 

Sample practically pure; theory, 18.36 per cent K ? S0 4 . 


58 CHEMICAL CALCULATIONS 

95. What is the percentage of copper carbonate, CuC0 3 , in 

a sample of malachite, Cu(OH) 2 . CuC0 3 , which analyzed for 57.1 
per cent copper? Ans. 55.49 per cent CuC0 3 . 

The salt is probably pure; theory, 55.87 per cent CuC0 3 . 

96. A sample of carnallite, KC1. MgCl 2 .6 H 2 0, analyzed for 

35.34 per cent chlorine. What is the percentage of magnesium 
chloride present? A ns. 31.64 per cent MgCl 2 . 

Note. —Theory: 34.27 per cent MgCl 2 . Therefore 92.31 per cent 
pure. 

97. A sample of crystallized ferrous-ammonium sulphate, 

FeS0 4 . (NH 4 ) 2 S0 4 .6 H 2 0, analyzed for 8.66 per cent ammonia, 
NH 3 . What is the percentage of ferrous sulphate, FeS0 4 , 
present? Ans. 38.60 per cent FeS0 4 . 

Theory, 38.74 per cent FeS0 4 . 

98. A sample of gahnite, Zn(A10 2 ) 2 or ZnO . A1 2 0 3 , weighing 

0.1909 gram, gave on analysis 1.664 grams of zinc sulphate, 
ZnS0 4 . What was the percentage of alumina, A1 2 0 3 , in the 
sample? Ans. 55.18 per cent. 

If analysis is correct, the sample is slightly impure. 

Theory, 55.67 per cent. 


CHAPTER VIII. 


THE DERIVATION OF CHEMICAL FORMULAE. 

Molecular Formulae from Molecular Weight and Per¬ 
centage Composition. — The formula of a compound, 
designated by symbols which stand individually for one 
atomic weight of an element, is derived from data afforded 
by analysis or synthesis. In reverse to the determination 
of the percentage composition, as described in the preced¬ 
ing chapter, we can derive the true molecular formula 
when given the correct molecular weight of a compound 
and the percentage amount of each element present. 
The proportional parts of the molecular weight indicated 
by the several percentages are of course the quantities 
(formula-quantities) of the respective elements in one 
molecular weight of the compound, and necessarily must 
be multiples of the corresponding atomic weights by unity 
or a small integer. 

Example 26. —The molecular weight of water is 18.016. The 
amount of hydrogen present is 11.2 per cent and that of oxygen 
88.8 per cent. What is the molecular formula? 




Proportional 

Unit or 


Molecular 

Percentage 

part of molec¬ 
ular weight 
corresponding 
to element. 

atomic 

Number of 

weight. 

indicated. 

weight of 
element. 

units. 

18.016 

H = 11.2% 

2.016 

1.008 

2 

18.016 

0= 88.8% 

16.00 

16.00 

1 


59 








60 


CHEMICAL CALCULATIONS 


The values 2.016 and 16 represent respectively the sum of 
the atomic weights of hydrogen and oxygen which are 
present in the molecular weight of water. They are, 
therefore, the formula-quantities of these elements in 
this compound. With 1.008 and 16 as the atomic weights 
of hydrogen and oxygen respectively, it is only a simple 
step to determine the number of hydrogen units (2) and 
oxygen units (1) which are necessary to make up the 
formula-quantity of each element in this molecule and 
consequently, together, the molecular weight of water; 
hence the formula H 2 0. 

The correct molecular weight of a compound is rarely 
ever at hand for the determination of chemical formulae. 
The usual procedure lies, then, in the determination of 
the formulae from the percentage composition or other 
data, and in the final adjustment of these to accord with 
the molecular weights which may have been determined 
only with approximate exactness. 

Empirical Formulae from Percentage Composition.— 
Upon the percentage basis the molecular weight is re¬ 
garded as brought over to the value 100; this number, 
however, gives in itself no clue to the probable molecular 
weight. In Example 26 we were given the correct molec¬ 
ular weight; consequently, the proportional parts of this 
total weight, as indicated by the percentage amounts, 
agree always with the exact quantity of each element 
present in one molecular weight of the compound. Now, 
upon the adoption of 100 as the molecular weight, each 
percentage amount becomes the accepted value for the 
formula-quantity of that particular element in the molec¬ 
ular weight. In dividing each of these values through by 
the atomic weight of the corresponding element, we obtain 
not whole numbers, indicating the number of atomic 
weights of the respective elements present, but fractions 


DERIVATION OF CHEMICAL FORMULAS 


61 


or factors of these whole numbers, all of which are 
related to the true numbers in the same ratio as the 
adopted value, 100, is related to the true molecular 
weight. 

Since a chemical formula calls for simple multiples of 
the atomic weights concerned, we need only raise the 
entire range of factors by some term which will bring 
each and all into whole numbers. When the smallest 
factor is made the divisor, then all of the other factors 
divided through by it must necessarily give quotients 
which are equal to or greater than this smallest factor as 
unity. In bringing the correct molecular weight over to 
100 as a basis, this smallest factor was of course reduced 
from unity, or a multiple of it, in the same proportion as 
were all of the other factors. The quotients, then, upon 
the basis of this small factor as unity, will possess values 
close to their former and correct numbers. If the smallest 
factor had been reduced from unity itself, then the quo¬ 
tients will represent the correct formula, — a molecular 
formula , — of the compound. If, however, the original 
value of this smallest factor was a simple multiple of unity 
( e.g ., 3), then the quotients will vary from their true values 
in a molecular formula by just this same fractional amount 
that unity is of the simple multiple. (If 3 were the mul¬ 
tiple, then the quotients would stand at one-third of their 
original values.) This simplest expression of a formula in 
symbols is known as an empirical formula , and the sum of 
the atomic weights therein represented, though a formula 
weight, is not necessarily the molecular weight. 

From the empirical formula the true percentage com¬ 
position of a compound is always derivable. Its formula 
weight, however, may not be coincident with the molecular 
weight, and, if so, the correct molecular formula can be 
derived only when the molecular weight, or at least a fair 


62 


CHEMICAL CALCULATIONS 


approach to this, is known. The molecular formula, there¬ 
fore, is always a multiple of the empirical formula by some 
small integer. 

Example 27. — A substance by analysis was found to contain 
32.32 per cent sodium, 22.44 per cent sulphur, and 45.24 per 
cent oxygen. What is the formula of the compound? 

The method of solution will be made most apparent 
when we set down the percentages and refer these to the 
respective atomic weights of the elements present, as 
indicated below: 


Substance. 

Percentage. 

Percentage 
referred to 
corresponding 
at. wt. as basis 

Factor. 

Factor to unit 
value. 

Sodium.. 
Sulphur.. 
Oxygen. . 

32.32 

22.44 

45.24 

100.00 

32.32/23.00 

22.44/32.07 

45.24/16 

1.406 ] Divide | 2.01 
0.699 [ through \ 1.00 
2.827 J by 0.699 l 4.04 


These unit values approach very closely to the integers 
2, 1 and 4, and designate the number of unit or atomic 
weights of sodium, sulphur and oxygen, respectively, in 
their corresponding formula-quantities in a formula 
weight of the compound, Na-jSC^. 

In Chapter YII we have noted the means for calculating 
the theoretical percentage composition of a compound. 
In the case of a derived formula it is always well to calcu¬ 
late from it the percentage amount of each element present 
and note whether or not these values check with those 
derived by analysis. The derived formula Na 2 S0 4 presents 
the following percentage composition: 












DERIVATION OF CHEMICAL FORMULAS 


68 


Na 2 S0 4 : 2 Na 
142.07 : 46 


| = 100 : x, or 32.38 per cent sodium. 
| = 100 : x , or 22.57 per cent sulphur. 
[ = 100 : x, or 45.05 per cent oxygen. 


Na 2 S0 4 : S 
142.07 : 32.07 


Na 2 S0 4 : 4 O 
142.07 : 64 


100 per cent. 


The percentage amounts actually found by analysis 
approach very closely to these theoretical values based 
upon the formula Na 2 S0 4 ; consequently we may conclude 
that this formula represents the constitution of the com¬ 
pound. The molecular weight of sodium sulphate is 
found to be close to 142; hence the formula Na 2 S0 4 
(with the formula weight 2 (23) + 32.07 + 4 (16) or 
142.07) satisfies also the requirements for a molecular 
formula. 

Accuracy in analytical data is dependent upon purity 
of material and method of operation. In all of our 
results we may expect a certain degree of variation from 
the theoretical values, but the limits of error in the deter¬ 
mination of chemical formulae should not greatly exceed 
two-tenths of 1 per cent. Much depends, however, upon 
the particular element considered. For example, in the 
case of hydrogen the error often may be as much as four- 
tenths of 1 per cent above the theoretical, due to insuffi¬ 
cient removal of moisture from the sample or to a faulty 
combustion. In the case of oxygen the data are rarely 
ever determined directly, but by difference; thus in 
Example 27 the sum of the percentage amounts for 
sodium and sulphur was subtracted from the total 100 
per cent, and the difference considered as the value for 
oxygen, a value altogether dependent upon the accuracy 



64 


CHEMICAL CALCULATIONS 


of the other data. From these considerations the chemist 
finds it greatly expedient, in problems of this nature, to 
consider each analysis separately and to determine what 
one is likely to be most free of error. The factor, there¬ 
fore, which corresponds to the percentage amount of this 
one element is the best to select as a basis for the reduction 
of all the other factors to unit values. Let us suppose, 
in Example 27, that the value for sodium was known to 
be more accurately determined. The factor here is 1.406, 
and, if we base all the other data upon this one, i.e. f 
by dividing each factor by 1.406, we shall obtain the values, 
1, 0.498 and 2.010, for sodium, sulphur and oxygen respec¬ 
tively. These may be brought to unit values by multi¬ 
plication by a small integer, — here by 2, — and we obtain 
finally 2, 0.996 and 4.02, — values which approach the 
respective integers 2, 1 and 4 somewhat more closely than 
in the previous calculation. This, of course, is due to the 
close agreement between the theoretical and found per¬ 
centage values for sodium. It is now seen why the factor 
for oxygen is rarely selected as a basis for formula deter¬ 
minations. 

Example 28. — Determine the formula for that substance 
which presented, by analysis, the following percentage com¬ 
position: carbon, 39.78 per cent, hydrogen, 6.97 per cent and 
oxygen, 53.25 per cent. Above 230° the vapor of this sub¬ 
stance gave a constant relative density, 2.09 (air = 1), calcu¬ 
lated to standard conditions. 

From the relative density it is only a simple step to 
calculate the molecular weight (c/. Ex. 14): 

1 : 2.09 = 28.955 : x, 

where x, the molecular weight, is found to be 60.52. 

The following table of percentages and corresponding 
atomic weights is easily constructed as under Example 27 : 





DERIVATION OF CHEMICAL PORMTJLA5 


65 


Substance. 

Percentage. 

Percentage 
referred to 
corresponding 
at. wt. as basis. 

Factor. 

Unit value. 

Carbon.... 

39.78 

39.78/12 

3.315 1 

1 Divide 

1.00 

Hydrogen . 

6.97 

6.97/1.01 

6.90 

[ through • 

2.08 

Oxygen.... 

53.25 

53.25/16 

3.33 J 

1 by 3.315 

1.01 


These unit values approach well the integers 1, 2 and 1, 
which represent now the number of atomic weights of these 
respective elements in a formula weight of the compound, 
CH 2 0. The sum of the formula-quantities thus repre¬ 
sented in the formula weight is 30.02 (i.e., 12 + 2.02 + 16), 
a value which does not coincide with the molecular weight 
calculated from the vapor-density determination above 
(60.52). The molecular formula as previously stated is 
always a multiple of this simplest or empirical formula, 
and by just that integer which brings the empirical formula 
weight up to the molecular weight or an approximation 
to the same. The integer in the present case is 60.52/30.02 
or 2.02. This is well within the limits of error, which in 
some cases may exceed a variation of 0.2 from an integral 
value. The empirical formula CH 2 0 must be multiplied 
now by the integer 2, when we shall obtain C 2 H 4 0 2 , the 
correct molecular formula, or that formula in which the 
sum of the atomic weights involved gives the molecular 
weight. The calculation of the theoretical percentage 
composition for this substance, acetic acid, gives the 
following: carbon, 39.97 per cent, hydrogen, 6.73 per cent 
and oxygen, 53.50 per cent. The analytical data, there¬ 
fore, are closely in accord with the theoretical. 

Formulae from Percentage Composition involving Radi¬ 
cals. — As a more complex example it will be well to 










66 


CHEMICAL CALCULATIONS 


examine the analytical data upon some mineral on which 
the results are given in percentage amounts of the several 
groups present. Feldspar belongs to this class of sub¬ 
stances, and in its analyses certain values will be found 
for silicon dioxide (Si0 2 ), aluminium oxide (A1 2 0 3 ) and 
potassium oxide (K 2 0); but associated with these there 
may occur other oxides which are isormorphous with one 
or more of the above. Consequently the factor for these 
isomorphous substances, found, of course, as heretofore 
described, by the relation of percentage composition to 
atomic weight or groups of atomic weights representing the 
substance, must be reckoned together as one factor and 
hence as one group, in order to determine the extent to 
which this particular group of isomorphous substances 
may be present in the complete formula. If, perchance, 
a portion of the data were not given in terms of the 
proper isomorphous compound, but in some part of it, for 
instance, the percentage of potassium instead of potassium 
oxide, we should require only a single calculation to con¬ 
vert the data into the correct form (c/. Ex. 22): 

2 K : K 2 0 = — per cent given : — per cent sought. 

Example 29. — A sample of the mineral orthoclase (a feld¬ 
spar) gave by analysis the following percentage composition: 
Si0 2 , 65.69 per cent, A1 2 0 3 , 17.97 per cent, (CaO, 1.34 per 
cent, K 2 0, 13.99 per cent, Na 2 0, 1.01 per cent, — isomor¬ 
phous). What is its formula? 

By drawing up a table of these percentages and referring 
each to the formula weight of the respective groups, we 
may expect to obtain factors which in this case do not 
refer to the simplest relation between the elements, but 
rather between the groups of elements. 

The factors for K 2 0, Na 2 0 and CaO, since these oxides 
are isomorphous and mutually may replace each other in 


DERIVATION OF CHEMICAL FORMULAE 


67 


the molecule, must be added together and considered as 
some generic formula such as R 2 0, when in the final 
adjustment of the number of each group present in the 
formula they will figure as an individual group of uniform 
composition. 


Sub¬ 

stance. 

Per¬ 

centage. 

Percentage re- 
referred to 
formula 
wt. of group 
as basis. 

Factor. 

Factor of isomor¬ 
phous groups. 

Unit 

value. 

Si0 2 . . 

65.69 

65.69/60.3 

1.0894 

1.0894 



6.258 

ai 2 o 3 . 

17.79 

17.79/102.2 

0.1741 

0.1741 

Divide 

1.000 

CaO.. 

1.34 

1.34/56.09 

0.0239] 


through• 


K 2 0.. 

13.99 

13.99/94.2 

0.1485 j- 

0.1887 

fly 


1.084 

Na 2 0. 

1.01 

1.01/62.2 

0.0163J 


0.1741 




As has been already noted in selecting the factor of 
those equally small, for division into the entire range of 
factors, it is always better to select that one which is 
likely to have been based upon more accurate analytical 
results, — in this case the aluminium oxide and not the 
composite oxide (into the determination of which three 
analyses must have entered). The resulting integers are 
here found to be 6, 1 and 1, and consequently the formula 
for the feldspar is (K 2 0 . CaO . Na 2 0) (A1 2 0 3 ) 6 (Si0 2 ), in 
which the oxides of the metals K, Na and Ca make up one 
single group of the general formula of a feldspar, — 
(K 2 0) . (A1 2 0 3 ) . 6 (Si0 2 ), or KAlSi 3 0 8 . The extent of 
these isomorphous replacements within any group may 
vary considerably. This empirical formula is accepted 
as the true formula in the absence of any conflicting 
statements in regard to the true molecular weight. 

Formulae from the Relation of Formula-Quantities to 
Each Other. — Without recourse to percentage composi- 












68 


CHEMICAL CALCULATIONS 


tion, the actual weight of any element found by analysis 
to be present in a known weight of compound may be 
referred directly to the formula-quantity of this element 
in the total molecular weight of that compound. Since 
these analytical data are dependent upon the number of 
atomic weights of each element present in one molecular 
weight, the ratio between the weight of each element in a 
known weight of compound containing, for example, two 
constituents, must be equal to the ratio between a certain 
unknown number of atomic weights of the one element 
and that of the other, i.e. f to the ratio between the cor¬ 
responding formula-quantities. 

Example 30. — Determine the formula of an iron oxide pro¬ 
duced in the oxidation of 22.4 grams of iron to a final weight of 
32 grams. 

Here 32 — 22.4 = 9.6 grams of oxygen taken up. The 
ratio between the weight of each of these two constituents 
in this sample of oxide, i.e., 22.4/9.6, must be equal to the 
ratio between the formula-quantity of each corresponding 
element in the molecular weight. 

Since the number of atomic weights of each element, 
grouped in its formula-quantity, is here unknown, only 
the algebraic expression Fe x O y can be written for the 
complete formula; x and y representing, of course, these 
unknown integers. Given, then, the atomic weights of 
these two elements, iron (55.85) and oxygen (16), we may 
easily draw up the ratio between the formula-quantity of 
each in the compound. This ratio involves an unknown 
factor (an integer) in each term, but as a ratio it must be 
always equal to any other ratio that can be drawn up 
from data on the weights of these same particular sub¬ 
stances in a definite weight of the compound. Thus the 
ratio Fe x /O y , as determined in the analysis of the oxide 
given above, is represented by 22.4/9.6, and consequently, 


DERIVATION OF CHEMICAL FORMULAE 69 

if the analyses are correct, these two ratios must be 
identical: 

Fe x = 22A 55.85 x = 22,4 

0„ ~ 9.6 ’ 0r 16 y 9.6 ’ 

The exact numerical relation between x and y may be 
readily calculated by bringing the numerical values to one 
side of the equation ( i.e ., by multiplying the equation 
through by 16/55.85), when we obtain 

x 22.4 X 16 358.4 

y ~ 9.6 X 55.85 “ 536.4' 

a result likewise obtained through the simple proportion: 

55.85 x : 16 y = 22.4 : 9.6 
(55.85 X 9.6) x = (16 X 22.4) y 
x/y = 358.4/536.2. 

In one equation involving two unknown terms we can 
only expect to determine the simplest ratio between them. 
The ratio 358.4/536.4 when reduced by division of each 
term by the smaller (358.4), gives us 1/1.49, and this in 
turn, through multiplication of each term by 2, is brought 
very close to 2/3, which, as the ratio between the smallest 
integers, indicates the value of x and y respectively; 
hence the formula of the compound, Fe 2 0 3 . 

Whether the molecular formula calls for this simplest 
form or some multiple of it by a simple integer cannot be 
determined without knowledge of the molecular weight 
of the compound. 

The application of this method to compounds contain¬ 
ing three or more elements introduces, of course, this 
corresponding number of unknown terms (the number 
of atomic weights of each element present), and lends 
itself less readily to a simple solution. When, however, 
two or more groups of elements, as radicals, etc., are 
present, the method may be applied to the determination 
of the number of each of the two groups in the molecule. 





70 


CHEMICAL CALCULATIONS 


Example SI. — What is the formula of that oxy-halogen salt 
of potassium, (KCl^O*, 13.9 grams of which lost 6.4 grams of 
oxygen upon heating, and gave a residue of 7.5 grams (13.9—6.4) 
of potassium chloride, KC1? 

The ratio between the weight of oxygen driven off and 
the residue of potassium chloride left is 6.4/7.5, which, of 
course, is equal to the ratio between the formula-quantity 
of each: 

O x = 6_4 or (16) x = 6,4 
(KC1)„ 7.5’ (39.1+35.46)y 7.5’ 

or 

x = (6.4) (74.56) = 476.2 
y “ (7.5) (16) ” 120 ' 

From this we obtain x/y = 3.97/1 or, by simplest integers, 
x = 4 and y= 1. The formula of the compound (KCl^O* 
now becomes (KC1) 1 0 4 , i.e., KC10 4 . 

Formulae Containing Water of Crystallization. — Again 
we have a good illustration of examples of this kind in 
the consideration of those salts which contain water of 
crystallization, — definite in amount under certain defi¬ 
nite and fixed conditions. 

Example 32. —7.15 grams of crystallized sodium carbonate, 
Na2C0 3 . x H 2 0, lost all of its water of crystallization when 
gently heated. The weight of the residue was 2.655 grams. 
What is the formula of the crystallized salt? 

Here 7.15 — 2.655 = 4.495 grams, or the weight of 
water in this weight of salt. The ratio between the weight 
of water actually associated with the weight of anhydrous 
salt (that portion of the crystallized salt left after the 
elimination of the water) is 4.495/2.655. This ratio there¬ 
fore must be equal to the ratio between the formula-quan¬ 
tity of water and the formula-quantity of the anhydrous 
salt in the molecular weight of the crystallized salt. In 
that the water of crystallization is usually determined 






DERIVATION OF CHEMICAL FORMULAS Tl 

with reference to one molecule of the anhydrous salt, one 
of the unknown terms in the ratio between the formula- 
quantities drops out (i.e., equals unity) and we have only 
the determination of the unknown number of molecular 
weights ( x ) of water present. Thus: 

x (H a O) 4.495 

Na 2 C0 3 2.655 * 

By substituting the atomic weights indicated, we obtain 

x (18.02) 4.495 

106 2.655' 

which, when solved for x, gives the value 10 as the number 
of molecules of water associated with one molecule of the 
anhydrous salt to form a molecule of crystallized sodium 
carbonate, Na 2 C0 3 .10 H 2 0. 

The calculation may be conducted by simple proportion 
as follows: 

Na 2 C0 3 : x (H 2 0) = wt. anhydrous salt : wt. H 2 0. 

106 : x (18.02) = 2.655 : 4.495. 

The expression x (18.02) is found equal to 179.5, hence the 
integral value for x will be 10 and accord with the formula 
above. 

Formulae from the Relation between Formula-Quan¬ 
tities and Their Corresponding Weights. — This constancy 
in the ratios between the quantities of all the elements 
either singly or collectively in a chemical compound was 
comprehended, of course, in the Law of Definite Pro¬ 
portions; and the proportions in which the elements enter 
into chemical combination are seen, accordingly, to be 
functions of the corresponding atomic weights. In any 
known amount of substance there is always a definite 
ratio between the actual weight of some one element, or 
group of elements, present and the corresponding formula- 






72 


CHEMICAL CALCULATIONS 


quantity. Furthermore, this ratio is always identical with 
every other ratio between the weight of any other element 
in this same sample and its corresponding formula-quan¬ 
tity.* The ratio can be unity only when we are dealing 
with a gram-molecular weight of the substance. Thus, 
in 18 grams of water, the weight of oxygen (16 grams) 
bears the ratio of unity to the formula-quantity of this 
element in the molecular weight, and also the same ratio, 
unity, exists between the weight of hydrogen (2 grams) 
and its formula-quantity. 

As is often the case, no one correct formula-quantity 
is at hand. Under such circumstances it is only possible 
to refer any and all of the known weights of the elements 
in a given sample of compound directly to their corre¬ 
sponding atomic weights. In order to maintain through¬ 
out the same constant ratio between these values — 
actual weights and atomic weights — it is necessary to 
select the ratio between the actual weight of some one 
element in the sample and the atomic weight correspond¬ 
ing thereto as the standard ratio. By means of this ratio 
the actual weight of any other element present in the 
sample may be brought over immediately (simple pro¬ 
portion) to a value which holds the same relation to its 
true formula-quantity as the single atomic weight of the 
first element will be found to have toward its formula- 

* The equivalence in the ratios between any two formula-quantities, 
as# and y, and the corresponding weights, as w l and w 2 , which repre¬ 
sent them respectively in any given sample, is usually expressed as 
x/y = w l /w r By multiplying through by y/w l we immediately 
obtain the expression x/w 1 = y/w 2 for this equivalence in the ratios 
between each formula-quantity and the weight representing it in some 
sample. The two expressions take the following forms through 
simple proportion: x : y = w x : w 2 and x:w l = y:w v each of which 
follows algebraically from the other, i.e ., the ratio between the first 
and second terms of a proportion, when equal to the ratio between the 
third and fourth terms, signifies also this equivalence in the ratios 
between the first and third terms and the second and fourth terms. 



DERIVATION OF CHEMICAL FORMULAS 


73 


quantity. In this latter case, with the formula-quantity 
always a multiple of the atomic weight by a small integer, 
the process of determining the true formula-quantity is 
comparatively a simple one. The integer selected, how¬ 
ever, must also bring each and all of the other values, 
similarly found, to multiples of their individual atomic 
weights. In other words, we determine the least common 
multiple for the entire range of values, such that each in 
order will be raised to a multiple, by unity or a small 
integer, of its atomic weight. Finally, the quotients of 
these so derived formula-quantities by the corresponding 
atomic weights will give the units of each element neces¬ 
sary in the simplest formula; one which can be raised 
later to any desired multiple, depending, of course, upon 
the molecular weight in question. 

Example S3. — Derive the formula of iron oxide from the data 
in Example 30 : 22.4 grams of iron gave 32 grams of oxide. 

The weight of oxygen entering into the compound is 
9.6 grams. Accordingly, the ratio 9.6/16, between weight 
of oxygen present in the sample and the atomic weight 
of this element, may be taken as the standard ratio. 
By means of this ratio the weight of iron present is now 
brought over to a value always equivalent to oxygen as 
16 in this compound.* Thus, 9.6 : 16 = 22.4 :x, where 

* In the combinations between hydrogen and oxygen no com¬ 
pound has been found to contain more than two atomic weights of 
hydrogen with one of oxygen in a single molecular weight. Con¬ 
sequently the lowest weight of oxygen corresponding to the lowest 
weight of hydrogen (1.008, its atomic weight) will be one-half of 
16, or 8. This weight of oxygen, the smallest equivalent to one 
atomic weight of hydrogen, is often used as a basis for the conver¬ 
sion of known weights or percentages of other elements (associated 
with oxygen) to the scale of atomic weights. Though it brings into 
the calculations numbers somewhat smaller (by one-half) than other¬ 
wise obtained through the use of the actual atomic weight, 16, the 
process of determining the least common multiple is in no way 
simplified. 


74 


CHEMICAL CALCULATIONS 


this value for iron is found to be 37.2. By referring these 
values now to the respective atomic weights we have: 


Sub¬ 

stance. 

Relative 

amount in 

known 
weight of 
compound. 

Value con¬ 
structed 

upon oxy¬ 
gen as 16. 

Multiple 

Value 

raised to 
multiple of 
correspond¬ 
ing at. wt. 

Atomic 

weight. 

Unit 

value. 

Iron.. . 
Oxygen 

22.4 

9.6 

37.2 

16 

3 

111.6 

48 

55.85 

16 

2 

3 


From this the formula of the compound is found to be Fe 2 0 ; 


Formulae from the Relation between Formula-Quantity 
and Percentage Composition. — In extending this method 
to compounds which contain a number of elements, it 
serves equally well if we employ, in place of actual weights, 
the percentage amounts of the elements, or groups of ele¬ 
ments, present. These values, of course, are as definitely 
related to each other as any other fractions of the total 
weight of a substance. The ratio between the percentage 
of any one element and its atomic weight suffices, through 
simple proportion, to bring over each of the other per¬ 
centages to the corresponding values for these elements 
in the compound. 

Example 34. — Derive the formula of sodium sulphate from 
the data given in Example 27. 


Sub¬ 

stance. 

Percent¬ 

age. 

Value 

con¬ 

structed 

upon oxy¬ 
gen as 16. 

Multiple 

Value 
raised to 
multiple of 
correspond¬ 
ing at. wt. 

Atomic 

weight. 

Unit 

value. 

Sodium.. . 

32.32 

11.43 


45.72 

23 

1.99 

Sulphur.. . 

22.44 

7.94 

4 

31.76 

32.07 

0.99 

Oxygen. .. 

45.24 

16 


64 

16 

4 


100.00 



































DERIVATION OF CHEMICAL FORMULAS 


75 


The value of oxygen is made equal to the atomic weight 
of this element, and the ratio 45.06 : 16 is used to reduce 
all of the other values proportionately, e.g .: 

with sulphur 45.24 : 16 = 22.44 : x, or x = 7.94 and 

with sodium 45.24 : 16 = 32.32 : x, or x = 11.43. 

These final values represent correctly the relative amounts 
of each element. One, oxygen, was made to coincide 
with its corresponding atomic weight, hence each of the 
other values will represent a weight of that respective ele¬ 
ment in this compound equivalent to oxygen as 16; and, 
as this is the unit value for oxygen, the entire range of 
values can be raised only through multiplication by simple 
integers until the lowest possible formula-quantity for 
each is obtained. The integer 4, found by trial, here 
raises all to values that are multiples of the corresponding 
atomic weights. These formula-quantities when divided 
by the respective atomic weights give at once the number 
of atomic weights of each element in the molecule, hence 
the formula Na 2 S0 4 . 

The Derivation of Formulae of a Known Type from a 
Single Analytical Value. — If it is known to what general 
class (oxides, chlorides, sulphates, etc.) a particular sub¬ 
stance of simple type (not involving replacements) belongs, 
its complete analysis is not necessary for the determination 
of the formula. For instance, the percentage of copper 
in a sample known to be a copper chloride will be suffi¬ 
cient to this end. This follows by reason of the fact that 
the percentage amount of any element found comes from 
the ratio between its formula-quantity in the molecule 
and the total molecular weight, — a ratio represented, 
we shall say, by the expression, —mol. wt./rc (at. wt.). 
Upon analysis certain definite numerical values fall to 
this ratio; in percentages, for example, we have the 


76 


CHEMICAL CALCULATIONS 


ratio 100/per cent element found. We thus form the 
equation: 

100 _ mol, wt. 

% element found ~ x (at. wt. element) 

Example 85. — A sample of a chloride of copper gave upon 
analysis 47.9 per cent copper. What is the formula of the com¬ 
pound? 

The ratio between the known and theoretical values 
now becomes 

100/47.9 = mol. wt./x (63.57), 

or 100 : 47.9 = mol. wt. : x (63.57). 

As x is always a small integer it may be neglected for 
the first calculation: 100 : 47.9 = mol. wt. : 63.57. From 
this we obtain a value 132.71, as the molecular weight, 
in which one atomic weight of copper must be present 
to the extent of 47.9 per cent. The difference, therefore, 
or 69.14 ( i.e ., 132.71 — 63.57 = 69.14), must represent the 
atomic weights of chlorine associated with this one atomic 
weight of copper. The quotient of 69.14 by 35.46, the 
atomic weight of chlorine, gives 1.95, or practically 2, and 
we at once draw up the formula CuCl 2 (63.57 + 2 (35.46) 
= 134.49), with the molecular weight of which the an¬ 
alytical data are well in agreement; hence the correct 
formula. 

If there had been obtained some fractional quantity as a 
quotient (in place of 2 above), this quantity, together with 
the value for copper as unity, may be raised by multiplica¬ 
tion to integral values which indicate the probable formula. 
This, of course, is identical with multiplying the molecular 
weight found by x (the smallest integrals in their order) 
until a value is obtained that can be made up of atomic 




DERIVATION OF CHEMICAL FORMULAE 


77 


weights without fractional parts. This try-out of a for¬ 
mula is nothing more than the reverse of the method 
outlined in Example 23, wherein was shown the means of 
determining the amount of any salt present in any one of 
its samples when given the percentage of some one element 
in the salt. In place of the molecular weight there given 
we have here only to set down, in order, the trial values 
for this molecular weight which correspond to the possible 
formula. Thus: 

I Percentage amount) f 100 ^ ( Formula-quantity) _ ( Molecular weight) 

of copper j ( ) \ copper j ’ ( calculated. ) 

47.9 : X = 63.57 : 

When the value calculated for x , from any trial molecular 
weight, approaches closely to 100 we are assured of the 
correct value for this molecular weight and consequently 
of the correct formula. 


PROBLEMS. 

99. What is the formula of that substance which gave, by 

analysis, 26.9 per cent sodium, 16.58 per cent nitrogen, 56.52 
per cent oxygen? Ans. NaN0 3 . 

100 . What is the formula of that substance which gave, by 

analysis, 26.6 per cent potassium, 35.22 per cent chromium, 
38.18 per cent oxygen? Ans. K 2 Cr 2 0 7 . 

101 . What is the formula of that substance which gave, by 

analysis, 24.65 per cent potassium, 34.85 per cent manganese, 
40.5 per cent oxygen? Ans. KMn0 4 . 

102 . Derive the formula of that substance with the observed 
i relative density 0.8853(0 = 1) and, by analysis, the composition: 

85.41 per cent carbon and 14.64 per cent hydrogen, dns. C 2 H 4 . 

103. Derive the formula of that substance with the observed 

density 1.189 and, by analysis, the composition: 92.1 per cent 
carbon, 7.85 per cent hydrogen. Ans. C 2 H 2 . 




78 


CHEMICAL CALCULATIONS 


Ans. 


104. Derive the formula of that compound of hydrogen and 
oxygen which gave, by analysis, 5.93 per cent hydrogen. A 
determination of its molecular weight gave the value 31.8. 

Ans. H 2 0 2 . 

105. Derive the formulae of that oxide of nitrogen which 
gave, by analysis, 30.4 per cent nitrogen. In the solid state it 
was found to have a molecular weight of 92.4, whereas the 
actual density of its vapor above 140° was only 2.013. 

Solid, N 2 0 4 . 
Gas, N0 2 . 

106. Derive the formula of that acid which gave, by analysis, 
26.5 per cent carbon, 2.2 per cent hydrogen and the rest oxygen. 
The molecular weight was determined approximately as 90.4. 

Ans. H 2 C 2 0 4 . 

107. Derive the formula of the mineral chalcopyrite, a speci¬ 

men of which gave, by analysis, the following percentage compo¬ 
sition: 34.40 per cent copper, 30.47 per cent iron and 35.87 
per cent sulphur. Ans. CuFeS 2 . 

108. Derive the formula of the mineral dolomite which gave, 
by analysis, the following percentage composition: 

CaO 
MgO 
C0 2 


2L23 per rent j ' somor P hoU3 
47.67 per cent 


100.27 


109 

lowing: 


Ans. (CaO,MgO) (C0 2 ) 
or (Ca,Mg) C0 3 . 

». By analysis a specimen of melanterite gave the fol 


FeO 

20.37 

per 

cent 

MgO 

4.60 

per 

cent 

so 3 

29.80 

per 

cent 

H 2 0 

45.07 

per 

cent 


99.84 




Derive the formula. 


Ans. (FeO,MgO) (S0 3 ) . 7 H 2 0 
or (Fe,Mg)S0 4 .7H 2 0. 




DERIVATION OF CHEMICAL FORMULAE 79 


110. By analysis a specimen of xenotime gave the follow¬ 
ing: 

P 2 0 5 32.45 per cent 

Y 2 0 3 54.13 per cent ] 

Ce 2 0 3 11.03 per cent \ isomorphous 

Fe 2 0 3 2.06 per cent J 

99.67 

Derive the formula. 

Ans. (Y 2 0 3 ,Ce 2 0 3 ,Fe 2 0 3 ) . (P 2 0 6 ) 
or ([Y,Ce,Fe] 2 0 3 ) (P 2 0 5 ) 

111. By analysis a specimen of columbite gave the following: 


Nb 2 0 5 47.05 per cent 

Ta 2 0 6 34.04 per cent 

Sn0 2 

FeO 

MnO 


isomorphous 



100.34 

Derive the formula. 


Ans. ([Fe,Mn,Ca,Sn]0) . ([Nb,Ta] 2 O s ). 


112. By analysis a specimen of garnet gave the following: 


Si0 3 39.09 per cent 



| isomorphous 


CaO 35.75 per cent J 

H 2 0 0.15 per cent (neglected) 


100.04 

Derive the formula. 


Ans. 3 ([Ca,Mg,Mn,Fe]0) . ([Fe,Al] 2 0 3 ) . 3 (Si0 2 ) 
General type, 3 RO . R 2 0 3 .3 Si0 2 
or R" 3 R'" 2 (Si0 4 ) 3 . 


113. Derive the formula of the oxide produced when 6.87 
grams of barium unite with 1.6 grams of oxygen. 

Ans. BaO a . 




80 


CHEMICAL CALCULATIONS 


114 . Derive the formula of the oxide formed in the com¬ 

bustion of 2.61 grams of aluminium with oxygen to a final 
weight of 5.01 grams. Ans. A1 2 0 3 . 

115 . Derive the formula of the oxide produced by the com¬ 
bustion of 43.45 grams of lead with 4.48 grams of oxygen. 

Ans. Pb 3 0 4 . 

116 . Derive the formula of the oxide produced by the burn¬ 

ing of 2.5 grams of phosphorus in oxygen to a final weight of 
5.7 grams. Ans. P 2 0 5 . 

117 . Derive the formula of the nitrate, 19.7 grams of which 
were prepared from 10.4 grams of bismuth. 

Ans. Bi(N0 3 ) 3 . 

118 . Derive the formula of the nitrate, 17 grams of which 

gave a residue of 13.8 grams of sodium nitrite, NaN0 2 , upon 
heating. Ans. NaN0 3 . 

119 . Derive the formula of that chlorate, 4.165 grams of 

which lost 1.3 grams of oxygen upon heating and gave a resi¬ 
due of barium chloride, BaCl 2 . Ans. Ba(C10 3 ) 2 . 

120. Derive the formula of mercuric cyanide, 5.4 grams of 
which lost 1.1 grams of cyanogen upon heating. 

Ans. HgC 2 N 2 . 

121. Derive the formula of the double salt of ammonium 
sulphate and copper sulphate, 4.12 grams of which lost 1.81 
grams of ammonium sulphate upon heating. 

Ans. (NH 4 ) 2 S0 4 . CuS0 4 . 

122. Derive the formula of crystallized sodium sulphate, 8.16 
grams of which lost 4.51 grams of water upon dehydration. 

Ans. Na 2 S0 4 .10 H 2 0. 

123 . Derive the formula of crystallized copper sulphate, 
7.84 grams of which lost 2.79 grams of water upon dehydration. 

Ans. CuS0 4 .5 H 2 0. 

124 . Derive the formula of crystallized aluminium sulphate, 
9.54 grams of which lost 4.61 grams of water upon dehydration. 

Ans. A1 2 (S0 4 ) 3 . 18 H 2 0. 

125 . Derive the formula of aluminium hydroxide, 4.75 grams 
of which lost 1.64 grams of water and left a residue of AL0 3 . 

Am. Al(OH) 3 . 


DERIVATION OF CHEMICAL FORMULAE 


81 


126 . Derive the formula of that nitrate which gave, by 

analysis, 62.45 per cent lead, 8.68 per cent nitrogen, 28.85 
per cent oxygen. Ans. Pb(N0 3 ) 2 . 

127 . Derive the formula of that substance which gave, by 

analysis, 52.02 per cent carbon, 13.2 per cent hydrogen, 34.68 
per cent oxygen. Ans. C 2 H 6 0. 

128 . Derive the formula of that acetate which gave, by analy¬ 

sis, 63.61 per cent lead, 14.62 per cent carbon, 1.98 per cent 
hydrogen, 19.79 per cent oxygen. Ans. Pb(C 2 H 3 0 2 ) 2 . 

129. A salt of mercury known to be a chloride analyzed for 
26.15 per cent chlorine. What is the formula? Ans. HgCl 2 . 

130 . A salt known to be a nitrate analyzed for 62.4 per cent 

lead. What is the formula? Ans. Pb(N0 3 ) 2 . 

131 . An oxide of iron gave, by analysis, 69.80 per cent iron. 

What is the formula? Ans. Fe 2 0 3 . 

132 . An oxide of barium gave, by analysis, 81 per cent 

barium. What is the formula? Ans. Ba0 2 . 

133 . Derive the formula of the crystallized salt which, by 

analysis, gave the percentage composition: 19.98 per cent iron, 
11.47 per cent sulphur, 5.24 per cent hydrogen, 63.31 per cent 
oxygen. 10 grams of this crystallized salt lost 4.5 grams of 
water upon dehydration. Ans. FeS0 4 .7 H 2 0. 

134 . The percentage composition of a certain salt is: 15.6 
per cent chromium, 14.41 per cent sulphur, 4.79 per cent hydro¬ 
gen, 65.2 per cent oxygen. 10 grams of the crystallized salt 
lost 4 grams of water upon dehydration and gave a residue of 
Cr 2 (S0 4 ) 3 . What is the formula? Ans. Cr 2 (S0 4 ) 3 . 15 H 2 0 

135 . The percentage composition of a certain salt is: 27.51 
per cent calcium, 22.15 per cent sulphur, 1.02 per cent hydrogen, 
49.32 per cent oxygen. 10 grams of this crystallized salt lost 
0.6 gram of water upon dehydration. What is the formula? 

Ans. (CaS0 4 ) 2 . H 2 0. 

136 . 2.5 grams of a crystallized salt known to be a sulphate 
of iron, and containing 20 per cent of iron, lost 1.13 grams of 
water upon dehydration. What is the formula? 

Ans. FeS0 4 .7 H a O. 


CHAPTER IX. 


CALCULATIONS DEPENDING UPON CHEMICAL 
EQUATIONS. 

In calculations which depend upon chemical reactions, 
the equations representing these reactions must first be 
constructed. In all chemical equations the number of 
atomic weights of any one element concerned remains a 
constant; the relative amount of each element, therefore, 
will be alike for both sides of the equation. Naturally 
the valence of each element in the reaction under consid¬ 
eration must be known, as upon these factors the balancing 
of equations is dependent. This valence or measure by 
which each atom of any element can enter into combina¬ 
tion determines, accordingly, the number of other atoms or 
groups of atoms necessary for consideration in any reac¬ 
tion. When, however, an element undergoes a change of 
valence the arrangement of these atom groupings must be 
made to accord with this change (cf. Chap. XII). The 
representation of all substances in the molecular form — 
constituting here a molecular equation — has the great 
advantage of indicating the corresponding volume rela¬ 
tions between the gaseous substances by reason of the like 
molecular volumes of all substances in the state of vapor. 

Reaction-Quantities. — The action of metallic sodium 
upon water is shown in the following equation: 

2 Na +2 H 2 0 = 2 NaOH + H 2 

2 mol. + 2 mol. = 2 mol. + 1 mol. 

2 (23.00)+ 2 (18.02) = 2 (40.01)+ 2.02 

46.00 + 36.04 = 80.02 + 2.02 

82.04 = 82.04 

82 


Relative 

parts 

by weight. 



CHEMICAL EQUATIONS 


83 


In this simple equation each of the quantities repre¬ 
sented is definite and bears that relation to every other 
quantity as is indicated by the corresponding group of 
atomic weights present in each. These quantities may 
be considered as the reaction-quantities , — atomic or 
molecular quantities definite for any given reaction and 
always proportional to each other, such that, when in¬ 
volved together, the ratio between them will be exactly 
equal to the ratio between the actual weights, of whatever 
denomination, which may represent them. This propor¬ 
tionality, relative, of course, to the construction of the 
equation itself, follows naturally from the Law of Definite 
Proportions. The reaction equation only indicates the 
apportionment of the various atomic or molecular group¬ 
ings for the several compounds possible under the observed 
conditions. The formula-quantities going to make up any 
number of molecules of a compound indicated, still bear a 
definite ratio to every other formula-quantity, with the 
result, of course, that the weight which represents any of 
these reaction-quantities in a given equation must also 
bear a similar and definite proportion to each other, no 
matter whether they are upon the same or opposite sides 
of the equation. The reaction-quantity of sodium in the 
equation above is represented by two molecular weights; 
the reaction-quantity of sodium hydroxide, corresponding 
to this value for sodium, is also represented by two molec¬ 
ular weights. As both involve the same quantity of 
sodium they may be regarded also as equivalent 
quantities. The ratio between them, 2 Na/2 NaOH, is 
more simply expressed in the unimolecular form, 
Na/NaOH. 

The reaction-quantity of hydrogen directly proportional 
to the reaction-quantity of sodium in this equation is rep¬ 
resented by one molecular weight, or H 2 . The ratio 


84 


CHEMICAL CALCULATIONS 


2 Na/H 2 is definite, therefore, for all possible values for 
these substances in this reaction. 

Example 36. — What weight of sodium hydroxide and of 
hydrogen can be procured by the action of 50 grams of sodium 
upon water? 

From the equation just studied, the ratio 2 Na/2 NaOH, 
or Na/NaOH, i.e., 23/40.01, is constant for all proportional 
amounts of these two substances. Therefore the ratio 
50 lx, between this known weight of the metal (50 grams) 
and its proportional value in hydroxide (#), must be 
equal to the ratio above (the related terms in these ratios 
are of course placed similarly), and we shall have 

23/40.01 = 50/a;. 

By simple proportion this may be expressed as 
23 : 40.01 = 50 : x. 

From these expressions the value of x (the weight of 
sodium hydroxide) is found to be 86.98 grams. 

In an exactly similar manner the ratio between the 
reaction-quantities of sodium and hydrogen, 2 Na/H 2 , or 
Na/H, or 23/1.01 may be made equal to the ratio between 
the known or unknown weights of these substances con¬ 
cerned in the reaction. Between the weight of sodium 
(50 grams) and its unknown proportional weight of 
hydrogen (#) we have the ratio 50/x; this is therefore 
immediately referred to the ratio between the correspond¬ 
ing proportional quantities above, Na/H or 23/1.01. 
Thus, 23/1.01 = 50/x, or 23 : 1.01 = 50 : x, from which 
the value for x is found to be 2.2 grams. 

In general we may state that when any substance is 
concerned in a chemical reaction, the amount of any other 
substance, resulting directly or indirectly through this 
reaction, bears to the former a definite ratio and one always 


CHEMICAL EQUATIONS 85 

equal to the ratio between the corresponding reaction- 
quantities of the substances in the equation. 

Example 87. — What weight of magnesium will be required 
for the liberation of 10 grams of hydrogen from water or an acid? 

We may first consider the decomposition of water by 
magnesium as shown in the following: 

Mg +H 2 0 = MgO +H 2 
1 mol. + 1 mol. = 1 mol.+ 1 mol. 

Parts by weight. {24.32 +18.02 = 40.32 + 2.02 

and again, the reaction of magnesium upon an acid such 
as hydrochloric acid: 

Mg +2 HC1 = MgCl 2 + H 2 
1 mol. + 2 mol. = 1 mol. + 1 mol. 
Parts by weight. {24.32 + 2 (36.47) = 95.24 + 2.02 

In each case we find that the molecule of magnesium — 
containing one atomic weight — displaces, and is equi¬ 
valent to, one molecule of hydrogen — containing two 
atomic weights. The quantities Mg and H 2 are there¬ 
fore directly proportional to each other in both equations, 
and the constant ratio Mg/H 2 , or 24.32/2.02, must be equal 
to the ratio between the corresponding weights of these 
elements here concerned, i.e., x/ 10. From the equation 
24.32/2.02 = x/ 10, or the simple proportion 24.32 : 2.02 
= x : 10, we calculate the value of x to be 120.4 grams 
( (magnesium). 

In the more complicated reactions the relations between 
the several quantities on the two sides of an equation are 
often difficult to ascertain. To say nothing of reversible 
1 actions and the possibilities for dependent or secondary 
[ reactions to take place between certain of the quantities, 
we have assumed and must continue to assume that the 
! reaction in question takes place along the lines indicated 



86 


CHEMICAL CALCULATIONS 


in that equation best substantiated by the facts under the 
observed conditions. 

Example 38. — What weight of sulphur dioxide, S0 2 , can be 
obtained by the action of 10 grams of copper upon concentrated 
sulphuric acid? 

Copper acts upon sulphuric acid (cone.), to give sulphur 
dioxide, water and copper sulphate. Notwithstanding 
the slight amount of cuprous sulphide formed here through 
a secondary reaction, we shall represent the action by a 
single equation: 

Cu +2 H 2 S0 4 = CuS0 4 + 2 H 2 0 + S0 2 
63.57 64.07 

The reaction-quantities, Cu and S0 2 , are directly pro¬ 
portional; the ratio between them, Cu/S0 2 , or 63.57/64.07, 
is constant for all amounts here involved, and conse¬ 
quently equal to the ratio, 10/x, representing these sub¬ 
stances respectively in this example. From the equation 
63.57/64.07 = 10/x the value of x is calculated as 10.08, 
the weight in grams of sulphur dioxide. 

Calculation of Volume Relations Introduced by Chem¬ 
ical Equations. — In the consideration of gaseous sub¬ 
stances formed in these reactions, their gram-molecular 
quantities may be referred at once to the gram-molec¬ 
ular volumes as a basis (c/. Ex. 16). For example, to 
recall the illustrations at the beginning of this chapter, 
two gram-molecular weights of sodium displace one gram- 
molecular weight of hydrogen, which under standard 
conditions occupies 22,400 c.c. Whatever be the weight 
of hydrogen evolved, the ratio between the G.M.W. and 
this weight will be identical with that between the G.M.V. 
and the volume of hydrogen corresponding to the known 
weight above (c/. Ex. 15). 


CHEMICAL EQUATIONS 


87 


Example 89. — Calculate the volume of hydrogen, at 10° and 
750 mm., that will be evolved by the action of 10 grams of zinc 
upon an acid. 

The nature or strength of the acid here is of no concern, 
for, so long as the zinc is used up, the proportional amount 
of hydrogen must be displaced, and from this weight of 
hydrogen displaced its volume may be easily calculated. 
With sulphuric acid the reaction may be represented as 
follows: 

Zn + H 2 S0 4 = ZnS0 4 + H 2 
65.37 + 98.09 = 161.44 + 2.02. 


The reaction-quantities Zn and H 2 establish the ratio 
Zn/H 2 , or 65.37/2.02, as constant for all proportional 
amounts of these elements; hence the equation 65.37/2.02 
= 10 lx. From this the value of x is calculated as 3.24 


grams. 

A gram-molecular weight of hydrogen, 2.02, occupies 
22,400 c.c. at standard conditions. As previously stated 
the volume occupied by any other weight of the gas will 
stand in the same proportion to this gram-molecular 
volume as does its weight to the gram-molecular weight. 
Thus the ratio 2.02/3.24, between the known weights, will 
be exactly equal to the ratio between the corresponding 
volumes, — 2.02/3.24 = 22,400 \x. By simple proportion 
we should have 2.02 : 3.24 = 22,400 : x or, with reference 
to the fractional part of the volume occupied by 2.02 grams 
of the gas, we know that 3.24 grams must occupy 3.24/2.02 

3.24 X 22,400 


times this known volume, i.e. f - 


2.02 


- or 35,929 c.c. 


The value of x , as here calculated, must be adjusted now 
to the observed conditions of temperature and pressure. 
For this we need but a moment’s glance to see that the 
corrected volume, x', will be expressed by the equation 


, _ ... Q9Q X (760) (283) 
* - (35,929) (7g0) (273) : 


or x r = 37,742 c.c. 




88 


CHEMICAL CALCULATIONS 


The reverse process — that of determining the amount 
of any substance which will give a definite volume of gas 
measured at certain definite conditions—is made clear 
when the weight of this gas is ascertained (c/. Ex. 17). 
So also the consideration of aqueous vapor, when present, 
brings into these calculations only those methods already 
outlined in previous examples. 

Calculations with Reference to Degree of Purity. — 
Oftentimes the purity of materials employed in chemical 
operations does not come up to the standard or 100 per 
cent. In such cases, where the deviation from the theo¬ 
retical is known, the results found must be recalculated 
with reference to the standard purity. For instance, if 
the magnesium considered in Example 37 had been of 
only 80 per cent purity (contaminated, we shall say, with 
20 per cent of inert or extraneous matter), then our 
results, wrongly based upon a valuation of 100 instead of 
the actual 80, must be raised in accordance with the ratio 
existing between the actual and theoretical value, i.e. f 
according to the ratio 80/100. The result in the exam¬ 
ple cited would be changed through the simple pro¬ 
portion: 120.4 : x = 80 : 100; or 150.5 grams would be the 
weight of magnesium (80 per cent) necessary to give 
10 grams of hydrogen. 

Conversely, if it were desired to ascertain what weight 
of hydrogen could be obtained from this weight (150.5 
grams) of magnesium (80 per cent pure), then the result 
upon the basis of 150.5 grams of pure magnesium would 
need to be lowered in the same proportion: 100 : 80. 
The same end is attained by bringing into the original 
calculation just 80 per cent of the weight of magnesium 
(80 per cent pure), i.e., 120.4 grams in this example. 

These calculations afford, therefore, a simple means for 
estimating the degree of purity of a definite weight of a 


CHEMICAL EQUATIONS 


89 


I 


substance when the amount of some constituent con¬ 
cerned in one of its reactions is referred to the correspond¬ 
ing theoretical value. 

Example 40- — A specimen of marble weighing 5 grams 
evolved 2.1 grams of carbon dioxide (corresponding to a theo¬ 
retical volume of 1162.5 c.c. at 20° and 750 mm.) when acted 
upon by an acid. Estimate the degree of purity, assuming that 
all of the carbonate was present as calcium carbonate. 


The reaction with hydrochloric acid is here given: 
CaC0 3 + 2 HC1 = CaCl 2 + H 2 0 + C0 2 . 
100.09 44 


From this equation the proportionality between the 
reaction-quantities CaC0 3 (100.09) and C0 2 (44) leads to 

the equation ——— = —> where x, 2.2 grams, is the weight 

X 

of carbon dioxide theoretically possible from this weight 
of pure calcium carbonate. The specimen is accordingly 
21 

only — , or 95.45 per cent, pure (2.2 :2.1 — 100 : x). Calcu- 

ZdZi 

lating from the standpoint of the carbon dioxide, we natu¬ 


rally reach the same value: 


44 


2.1 


100.09 x 


or x, 4.772 


grams, is the theoretical amount of calcium carbonate re¬ 
quired for the production of 2.1 grams of carbon dioxide. 
4772 

This weight is — - -, or 95.45 per cent, of the weight of the 
ouuu 

specimen; consequently, just this fraction of the specimen 
can be considered calcium carbonate, i.e., its degree of 
purity is 95.45 per cent. 

Calculation of Products Resulting from Mixtures.— 

The definite weights of various substances brought 
together for chemical combination are rarely ever present 
in the exact proportions indicated by the equation 
representing the particular reaction. 





90 


CHEMICAL CALCULATIONS 


Thus, in the interaction between a metal and an acid, 
as already noted, the acid is taken in excess of the theo¬ 
retical amount proportional to the definite weight of 
metal concerned. Such an excess is easily removed 
from the final product by volatilization. In cases where 
this cannot be accomplished without decomposition of 
some product sought, other means (crystallization, etc.) 
are employed. In order to ascertain the exact amount 
of any substance formed through one of these inter¬ 
actions it is necessary first to determine what particular 
component or components can be acted upon to com¬ 
pletion by the others present. This, of course, presupposes 
the tendency for the reaction to run to completeness in 
some one direction, and implies the elimination of those 
products which may lead to a reversal of these conditions. 

Example — What weight of sodium sulphate may be 
expected to result from the interaction of 10 grams of sodium, 
11 grams of sulphur and 40 grams of oxygen? 

The equation representing the possible combination of 
these elements to this end is as follows: 

2 Na + S + 2 0 2 = Na 2 S0 4 
46 + 32.07 + 64 = 142.07 

Weights given. |10 11 40. 

The actual weights of the several substances here 
entering into combination as sodium sulphate must 
stand in the same relation to each other as do the cor¬ 
responding reaction-quantities. There is then an equiva¬ 
lence in the ratios between each reaction-quantity and the 
corresponding weight that represents it in the reaction 
(see footnote, page 72). 

If these conditions are fulfilled in the example, we 
shall have an equality in the ratios, 46/10, 32.07/11 
and 64/40. A single glance, however, disproves this 


CHEMICAL EQUATIONS 


91 


point. It then becomes necessary to determine by trial 
which one of these ratios, between a reaction-quantity 
and its corresponding weight as given in the example, is 
the correct ratio to select as the basis of calculation. For 
example, upon the basis of 64/40, we must have: 

— = — = , where x, (28.7), and y, (20), as the re- 

40 a: y 

spective values for sodium and sulphur, actually exceed 
the amounts at hand. Upon the basis 32.07/11, we have: 

" — = — = — > where x, (15.7), and y } (22), are the 
11 x y 

respective values for sodium and oxygen. In this latter 
case only the sodium is higher than the weight stipu¬ 
lated. Finally, upon the basis 46/10, we have the equa¬ 
tion — = ^ — = —} and from this both x, (6.97), the 
10 x y 

weight of sulphur, and y , (13.9), the weight of oxygen, 
fall below the amounts given; hence 46/10 is the proper 
ratio upon which to base the calculation. The sum of 
the respective weights thus obtained, 30.87 grams, (10 + 
6.97 + 13.97), gives the highest weight of sodium sulphate 
possible from the data in the problem. 

In general we select as the basis of calculation for any 
given reaction only that ratio which brings into consid¬ 
eration those values for the various substances involved 
as do not exceed the amounts present. More simply 
stated, perhaps, we base our calculations upon that weight 
of a particular substance at hand which gives, with its 
corresponding reaction-quantity as denominator, the 
smallest numerical factor. Thus the total weight of this 
particular substance determines the theoretically propor¬ 
tional weights of the other substances. 

Under conditions where reversible reactions are likely, 
for example the preparation of sodium sulphate by the 





92 


CHEMICAL CALCULATIONS 


action of sulphuric acid upon common salt, as shown in 
the equation: 

2 NaCl + H 2 S0 4 = Na 2 S0 4 + 2 HC1, 
we determine in similar manner the weight of sodium 
sulphate possible from any known weight of salt (if the sul¬ 
phuric acid is in excess) or from the weight of sulphuric 
acid (if the salt is in excess). The presence of hydro¬ 
chloric acid as the reversing agent must be removed in 
either case if we wish to obtain the calculated results. 

Calculation of Ratios between Reaction-Quantities in 
Dependent Equations. — When an element or group of 
elements enters into a series of successive reactions and 
the equation for each reaction can be constructed, we 
may draw up a proportionality between the reaction- 
quantities in any two of the equations providing that 
some quantity is common to both. In like manner this 
proportionality may be extended step by step over any 
number of dependent equations (c/. Ex. 24). 

Example 42. — What weight of bromine can be liberated 
from a concentrated solution of potassium bromide (excess) 
by the addition of 12 grams of hydrochloric acid (containing 
39.1 per cent HC1) and an excess of manganese dioxide? 

The two equations representing the action are as fol¬ 
lows: 

(а) Mn0 2 + 4 HC1 = MnCl 2 + Cl 2 + 2 H 2 0 

(б) 2 KBr + Cl 2 =2 KC1 + Br 2 . 

It is observed that the reaction-quantity Cl 2 is common 
to both (a) and (6), consequently the reaction-quantities 
directly proportional to this quantity in either equation 
will be also directly proportional to each other. The 
ratio 4HC1/C1 2 from equation (a) and the ratio Cl 2 /Br 2 
from equation ( b ) give us, accordingly, the ratio 4 HCl/Br 2 
for the proportionality between the reaction-quantities 


CHEMICAL EQUATIONS 


93 


required for this example. From the equivalence between 
this ratio and the ratio of the weights corresponding 
thereto, we have 4 HCl/Br 2 , or 4(36.47)/2(79.92), or 
145.88/159.84 = 12 /x, which, solved for x, gives a value 
of 13.1 grams. This weight of bromine, 13.1 grams, is 
based upon the hydrochloric acid as 100 per cent hydro¬ 
gen chloride. The concentrated acid at our command, 
the acid stated in the example, contained only 39.1 per 
cent hydrogen chloride. It remains then to calculate the 
weight of bromine which a 39.1 per cent acid can give. 
By reference to Example 40, the method is outlined to 
be simply one of proportion, according to which we shall 
have 100 : 39.1 = 13.1 : x, 

or x = 5.13 grams, the weight of bromine evolved by 12 
grams of 39.1 per cent hydrochloric acid. This same 
result is easily obtained by determining first the weight 
of hydrogen chloride in the 12 grams of 39.1 per cent acid, 
(12 X 39.1 per cent = 4.69), and working the example 
with this value for the hydrogen chloride. 

The elimination of a quantity occurring in two depend¬ 
ent equations may necessitate a readjustment of one or 
both of the equations containing it before the quantity 
becomes alike in each. 

Example 43. — What weight of iron sulphide will be required 
to furnish sufficient hydrogen sulphide for the reduction of 10 
grams of sulphur dioxide to sulphur? 

The two equations here required are as follows : 

(а) FeS + 2 HC1 = FeCl 2 + H 2 S 

(б) 2 H 2 S + S0 2 =2 H 2 0 + 3 S. 

The reaction in equation ( b ) depends upon the hydro¬ 
gen sulphide that is evolved in (a); consequently this sub¬ 
stance must constitute the common reaction-quantity. 
In order to make this quantity alike for the two equations 


94 


CHEMICAL CALCULATIONS 


and thus eliminate it from the calculations, it is only neces¬ 
sary to multiply equation (a) by 2, when we obtain (a'): 

(a 7 ) 2 FeS + 4 HC1 = 2 FeCl 2 + 2 H 2 S. 

The comparison of any of the reaction-quantities in the 
two equations (a 7 ) and (6) with reference to hydrogen sul¬ 
phide is now made simple. In the example given, iron 
sulphide and sulphur dioxide are found to be related 
through the ratios 2 FeS/2 H 2 S and 2 H 2 S/S0 2 , or directly 
as 2 FeS/S0 2 . The calculation is conducted, therefore, 
as follows: 

2 FeS/S0 2 , or 2 (87.92)/64.07, or 175.84/64.07 = x/ 10. 

From which x, the weight of iron sulphide, is found to be 
27.44 grams. 

Without this elimination of the quantity common to both 
equations, the calculation, of course, can be made directly 
toward ascertaining the weight representing this quan¬ 
tity in the first equation, and then finally, from this weight, 
the weight representing any other reaction-quantity pro¬ 
portional to it in the second equation. Such calculations, 
here involving the weight of hydrogen sulphide, are indeed 
roundabout and entirely unnecessary. 

Calculation of Ratios between Reaction-Quantities in 
Independent Equations. — In the study of dependent 
equations the reaction-quantities may be regarded as 
related to each other, through this common reaction-quan¬ 
tity, as are the members of a single equation. The reac¬ 
tion-quantities of the second equation are, so to speak, 
brought into existence through the agency of this com¬ 
mon quantity. In the study of independent equations, 
where there is present no one quantity which has a direct 
bearing upon any other equation, we have simply a fur¬ 
ther application of this same principle; namely, the com¬ 
parison of all the possible reaction-quantities in the 


CHEMICAL EQUATIONS 


95 


separate equations so long as some quantity can be made 
common to each. Quantities so compared will be directly 
proportional to each other, but only in respect to this 
common reaction-quantity. As an illustration, the fol¬ 
lowing equations are cited; 

(a) Na 2 C0 3 + 2 HC1 = 2 NaCl + H 2 0 + C0 2 

( b ) NaHCOg + HC1 = NaCl + H 2 0 + C0 2 . 

In these independent equations there are a number of 
substances represented for which the reaction-quantities 
could be adjusted alike for both; thus the reaction-quan¬ 
tity C0 2 is a common one. Upon this fact we may draw 
up the ratio Na 2 C0 3 /NaHC0 3 to express the proportion¬ 
ality between the relative amounts of normal carbonate and 
primary carbonate necessary to give an equal amount of 
carbon dioxide, i.e., the relation is based upon the carbon- 
dioxide content. In order to obtain a comparison with 
reference to the salt, equation (6) must be doubled to 
(&'): 

(i b ') 2 NaHC0 3 + 2 HC1 = 2 NaCl + 2 H 2 0 + 2 C0 2 . 

The ratio Na 2 C0 3 /2 NaHC0 3 then expresses the relation 
between the relative amounts of each carbonate neces¬ 
sary to give equal amounts of salt with hydrochloric acid. 
Or, since sodium is always a constant quantity in salt, 
we may say that the ratio above is that based upon a like 
content of sodium in each carbonate. 

In addition to the carbonate discussed in the preceding 
paragraph, we may also compare the reaction-quantities 
for the carbon dioxide present. Between equations (a) 
and (6), where this is a common quantity, we have the 
ratio 2 HC1/HC1 representing the relative amounts of acid 
necessary in the respective cases to give equal amounts 
of carbon dioxide. Between equations (a) and ( b ') we 
have the ratio C0 2 /2 C0 2 representing the relative 


96 


CHEMICAL CALCULATIONS 


amounts of carbon dioxide evolved from equal amounts 
of sodium when contained respectively in normal car¬ 
bonate or primary carbonate, as indicated by the ratio 
Na 2 C0 3 /2 NaHCOg, or the ratio 2 NaCl/2 NaCl, in each of 
which the sodium content is the same for both terms of 
the ratio. This latter point may be illustrated by the 
equation: 

2 NaHCO s = Na 2 C0 3 + C0 2 + H 2 0. 

Here it is shown that two molecules of the primary car¬ 
bonate break down into one molecule of the normal car¬ 
bonate with the loss of one molecule of carbon dioxide. 
The ratio 2 NaHC0 3 /C0 2 represents the relative amounts 
of these substances concerned in the action of heat upon 
the primary carbonate. This, in fact, is a case where the 
entire quantity of a substance, (C0 2 ), available in a com¬ 
pound need not be concerned in a ratio for the study of 
that compound, but only that portion of it separately in¬ 
volved as a reaction-quantity and made directly propor¬ 
tional, therefore, to some other reaction-quantity. 

Example 44- — What relative weights of mercuric oxide and 
barium peroxide are required in the preparation of equal 
amounts of oxygen? 

The molecular equations with oxygen as the common 
quantity adjusted alike in both are as follows: 

2 HgO = 2 Hg + 0 2 
2 Ba0 2 = 2 BaO + 0 2 . 

The ratio 2Hg0/2Ba0 2 , or HgO/Ba0 2 , or 216/169.37, 
determines accordingly the relation between the corres¬ 
ponding weights of these two substances in this problem. 
If we take one, e.g., the mercuric oxide, as 100 we re¬ 
duce the second to a comparatively simple value: 

216 _ 100 _ 

169.37 " x 



CHEMICAL EQUATIONS 


97 


This gives 78.4 as the value of x, or that weight in 
grams of barium peroxide equal to 100 grams of mercuric 
oxide in the preparation of oxygen. 

Calculation of Reaction-Quantities from the Weights 
of Substances Involved. — The direct proportionality 
which exists between the reaction-quantities of a given 
equation requires, as we have seen, the same proportion¬ 
ality between the corresponding weights which represent 
them in this particular reaction. This permits of an 
equivalence in the ratios between each reaction-quantity 
and its corresponding weight (c/*. Ex. 41 and footnote, 
page 72). 

In the construction of chemical equations from the 
actual weights of substances therein concerned, we must 
bear in mind the possibility of deviations in these weights 
from those demanded in the reactions. These deviations 
may be due to any number of causes and rapidly increase 
with the instability of the compounds considered as well 
as with the tendencies for secondary reactions. Conse¬ 
quently the determination of the correct reaction-quan¬ 
tities for these equations must be made a special study in 
each individual case. 

Example 45. — A solution of 10 grams of crystallized sodium 
thiosulphate, Na 2 S 2 0 3 .5 H 2 0 decolorized 5.1 grams of iodine. 
What molecular quantity of this salt was associated with one 
molecule of iodine in this reaction? 

From the equivalence in ratios between.the reaction- 
quantities and the actual weights involved wejiave: 

x I 2 x 253.84 

lo = 5l ’ or io ST - ' 

This gives to x the value 497.7, the reaction-quantity of 
the thiosulphate associated with one molecular weight of 



98 


CHEMICAL CALCULATIONS 


iodine. The molecular weight of sodium thiosulphate, 
Na 2 S 2 0 3 .5 H 2 0, is 248.2, consequently we have here 
497.7/248.2 or approximately 2 molecules of this salt in 
its reaction-quantity, i.e., the equation constructed upon 
the iodine involved as just one molecule will be 

2 Na 2 S 2 0 3 .5 H 2 0 + I 2 = (2 Nal + Na 2 S 4 0 6 ). 

Complex Reaction-Quantities. — There are a number of 
equations in which the reaction-quantities upon one side 
are incorporated into one single reaction-quantity upon 
the other side. The same principles hold here as in the 
cases just discussed, but the study of this larger quantity 
is nothing more or less than the study of the formula- 
quantities present in it. The consideration of the so-called 
molecular compounds, as are the double salts and salts 
containing water of crystallization, illustrates this point. 

Example 4-6. — 100 grams of copper sulphate will give what 
weight of blue vitriol (CuS0 4 .5 H 2 0) ? 

From the equation, — 

CuS0 4 + 5 H 2 0 = CuS0 4 .5 H 2 0 
159.64 + 90.1 = 249.74, 

the ratio 

CuS0 4 159.64 

CuSO,. 5 H 2 0 ’ ° r 249.74 

is a constant, and denotes the direct proportionality 
between these two reaction-quantities. The ratio between 
the actual weights, 100 and x, when put equal to the ratio 
between the molecular quantities, gives us 
159.64 = 100 
249.74 x ' 

where x, with the value 156.4 grams, is the weight of blue 
vitriol. As the percentage composition of the anhydrous 
copper sulphate is a constant, we may just as well calcu¬ 
late what weight of pure copper, Cu, or sulphur, S, or even 





CHEMICAL EQUATIONS 


99 


oxygen, 4 0, is necessary to give any definite weight of 
anhydrous copper sulphate and finally blue vitriol; or, 
vice versa, what weight of blue vitriol is obtainable from 
any definite weight of one of these constituents. 

As another illustration of this point we may take the 
formation of ordinary alum: 

KaSCh . 6H 2 0 + A1 2 (S0 4 ) 3 . 18H 2 0 = KaSO, . A1 2 (S0 4 ) 3 .241^0 
1 mol. + 1 mol. = 1 mol. 


A molecular quantity of one sulphate unites with one of 
another to form one molecular quantity of the double 
sulphate. These quantities are all directly proportional 
to each other, and from the equation we write the following 
ratios: 

, v K 2 S0 4 .6 H 2 0 

W K 2 S0 4 . A1 2 (S0 4 ) 3 .24 H 2 0 
A1,(S0 4 ), ■ 18 H 2 0 
W K 2 S0 4 . A1 2 (S0 4 ) 3 .24 H 2 0 
, K 2 S0 4 .6 H ? 0 

K ) A1 2 (S0 4 ) 3 . 18 H 2 0 ’ 

These serve as the ratios for calculating, upon the molecu¬ 
lar quantities involved, the amount of alum obtainable 
from certain known amounts of (a) crystallized potassium 
sulphate; (6) crystallized aluminium sulphate; and also 
(c) the amount of crystallized aluminium sulphate neces¬ 
sary for combination with one molecule of the crystallized 
potassium sulphate, or vice versa. Though somewhat 
more complicated molecular aggregates may be present, 
the ratios between the several factors are always constant. 


Example 47. — How much alum can be prepared from 100 
grams of anhydrous potassium sulphate? 


The ratio 

k^o; 


K 2 S0 4 

A1 2 (S0 4 ) 3 .24 H 2 0 


174.27 
0r 949.16 







100 


CHEMICAL CALCULATIONS 


is placed equal to the ratio 100 /x, and the equation solved 
in the usual manner. The value of x, the weight of alum, 
is found to be 544.6 grams. 

PROBLEMS. 

137. What weight of potassium hydroxide may be prepared 
by the action of 100 grams of potassium upon water? 

Ans. 143.5 grams. 

138. What weight of potassium will be required in the prep¬ 
aration of 20 grams of potassium carbonate, K 2 C0 3 ? 

Ans. 11.3 grams. 

139. What weight of magnesium chloride, MgCl 2 , may be 

obtained by the action of hydrochloric acid upon 10 grams of 
magnesium carbonate, MgC0 3 ? What weight of carbon dioxide 
will be liberated? Ans. 11.3 grams MgCl a . 

5.2 grams C0 2 . 

140. Recalculate Problem 139 on the supposition that the 
magnesium carbonate contained 10 per cent of insoluble matter. 

Ans. 10.16 grams MgCl a . 

4.7 grams CO a . 

141. What weight of sulphur dioxide, SO a , can be obtained 

by the action of an acid upon 250 grams of sodium sulphite, 
Na»S0 3 ? Ans. 127 grams. 

142. Recalculate Problem 141 on the supposition that 20 
per cent of the sulphite had become oxidized to sulphate. 

Ans. 101.6 grams. 

143. Calculate the volume of carbon dioxide, at 22° and 740 
mm. pressure, that will be liberated by the action of acid 
upon 20 grams of calcium carbonate, CaC0 3 . Ans. 4967 c.c. 

144. What weight of magnesium will be required for the 

liberation of 500 c.c. of hydrogen, at 20° and 740 mm. pressure, 
when acted upon by an acid? Ans. 0.49 gram. 

145. Calculate the volume of hydrogen, measured over water 
at 17° and 742 mm. pressure, that can be liberated by the 
action of 10 grams of sodium upon water. Ans. 5403 c.c. 

146. What weight of aluminium will be required for the 
liberation of 420 c.c. of hydrogen, measured over water at 18° 


CHEMICAL EQUATIONS 


101 


and 746.4 mm. pressure, when acted upon by hydrochloric 
acid? Ans. 0.306 gram. 

147. What weight of ammonium nitrite, NH 4 N0 2 , will 

evolve, when heated, 480 c.c. of nitrogen, measured over water 
at 21° and 747.5 mm. pressure? Ans. 1.22 grams. 

148. Determine the purity of a sample of anhydrous sodium 
carbonate, Na 2 C0 3 , 6 grams of which gave, when acted upon by 
an acid, 1310 c.c. of carbon dioxide, at 10° and 750 mm. pressure. 

Ans. 98.35 per cent pure. 

149. Determine the purity of a sample of anhydrous sodium 
carbonate, 3 grams of which gave, by treatment with sulphuric 
acid and final ignition, 3.99 grams of sodium sulphate, Na 2 S0 4 . 

Ans. 99.25 per cent pure. 

150. A sample of iron wire weighing 2.4 grams was found to 
give, when acted upon by excess of acid, a volume of hydrogen 
measuring 1015.1 c.c., at 10° and 745 mm. pressure. What was 
its degree of purity? The presence of any other substance 
capable of liberating hydrogen from an acid is here disregarded. 

Ans. 99.72 per cent pure. 

151. A sample of silver nitrate weighing 2.40 grams was 
brought into solution and treated with a soluble chloride 
(excess). The weight of silver chloride, AgCl, precipitated was 
2.01 grams. What was the purity of the sample? 

Ans. 99.26 per cent pure. 

152. What weight of zinc (98 per cent pure) will be required 
for the liberation of the hydrogen from 10 grams of hydro¬ 
chloric acid containing 39.1 per cent HC1? Ans. 3.576 grams. 

153. What weight of sulphuric acid containing 27.32 per 

cent H 2 S0 4 will be required for interaction with 2.17 grams of 
iron wire (99 per cent pure) ? Ans. 13.82 grams. 

154. What volume of hydrogen, measured over water at 18° 

and 746.4 mm. pressure, will be liberated by the action of 
aluminium upon 20 grams of sulphuric acid containing 41.5 
per cent H 2 S0 4 ? . Ans. 2100 e c. 

155. What weight of sulphuric acid (27.32 per cent H 2 S0 4 ) 

will be required for interaction with a metal (Zn, Mg, etc.) in 
order to give a volume of hydrogen measuring over water 1033 
c.c., at 16° and 742.5 mm. pressure? Ans. 15 grams. 


102 


CHEMICAL CALCULATIONS 


156. What weight of hydrochloric acid (23.82 per cent HC1) 
will be required for interaction with iron sulphide, FeS, in order 
to give a volume of hydrogen sulphide, measuring 613.1 c.c., at 
20° and 745 mm. pressure? 

Calculate also the weight of iron sulphide (98 per cent pure) 
here consumed. Ans. 7.655 grams acid. 

2.243 grams FeS (98 per cent). 

157. When 100 grams of mercury and 20 grams of sulphur 

are rubbed together what weight of mercuric sulphide, HgS, 
may be formed? Ans. 116 grams. 

Note. — From the reaction-quantities involved it is seen that the 
sulphur is in excess. This excess is easily removed by solution in 
carbon disulphide. 

158. A mixture of 10 grams of zinc dust and 2 grams of 
sulphur was gently heated to point of reaction. What weight 
of zinc sulphide, ZnS, was formed? Ans. 6.053 grams. 

159. A mixture of 10 grams of iron and 8 grams of sulphur 
was gently heated to point of reaction. What weight and 
volume (at standard conditions) of hydrogen sulphide could be 
obtained from the final product, FeS, by the action of an acid? 

Ans. 6.104 grams, or 4010.5 c.c. H 2 S. 

160. A mixture of 4 grams of sodium oxide, Na 2 0, and 6 

grams of sulphur trioxide will give what weight of sodium 
sulphate? Ans. 9.166 grams. 

161. 3 grams of silver nitrate, AgN0 3 , and 1 gram of potas¬ 
sium chloride, KC1, were brought together in aqueous solution. 
What weight of silver chloride, AgCl, was precipitated? 

Ans. 1.923 grams. 

162. 8.2 grams of crystallized barium chloride, BaCl 2 .6 H 2 0, 

and 7 grams of sulphuric acid (70 per cent H 2 S0 4 ) were brought 
together in aqueous solution. What weight of barium sulphate, 
BaS0 4 , was precipitated? Ans. 6.05 grams. 

163. 10 grams of a mixture of marble (CaC0 3 ), magnesium, 
and an inert substance were acted upon by excess of acid. The 
carbon dioxide evolved (taken up in a solution of potassium 
hydroxide) was found to weigh 1.318 grams. The volume of 
the other gas, hydrogen, measured over water at 10° and 


CHEMICAL EQUATIONS 103 

750.2 mm. pressure, was 5874 c.c. Calculate the percentage 
amount of each component in the mixture. 

Ans. 30 per cent marble. 

60 per cent magnesium. 

10 per cent insoluble matter. 

164. 15 grams of an alloy of zinc and copper (containing 

10 per cent of copper) were placed in a vessel containing 100 
grams of sulphuric acid (25 per cent H 2 S0 4 ). What weight of 
hydrogen was liberated? Ans. 0.417 gram. 

165. 12 grams of an alloy of aluminium and zinc (containing 
33J per cent of zinc) were placed in a vessel containing 180 
grams of hydrochloric acid (35 per cent HC1). What volume of 
hydrogen, at standard conditions, was liberated? 

Ans. 11,290 c.c. 

166. A specimen of silver, containing 3 per cent copper, 

weighed 9.8 grams. After solution in nitric acid an excess of 
sodium chloride was added to it. Calculate the weight of the 
silver chloride precipitated. Ans. 12.632 grams. 

167. 10 grams of phosphorus tribromide, PBr 3 , were mixed 

with an excess of water and an excess of silver nitrate added to 
this solution. What weight of silver bromide was formed? 
Calculate also the exact weight of silver nitrate required for 
this removal of bromide. Ans. 20.81 grams AgBr. 

18.82 grams AgN0 3 . 

168. What volume of hydrogen sulphide, at standard con¬ 
ditions, would be required for interaction with an excess of 
iodine, in aqueous suspension, in order to furnish an amount 
of hydriodic acid sufficient for the precipitation of 10 grams of 
silver iodide from a solution of silver nitrate? Ans. 477 c.c. 

169. What weight of fluorspar, CaF 2 , would be required to 

furnish sufficient hydrogen fluoride (by interaction with sul- 
iphuric acid) to convert 5 grams of quartz, Si0 2 , into silicon 
fluoride, SiF 4 ? Ans. 12.95 grams. 

170. Calculate the volume of chlorine, at standard condi¬ 

tions, necessary to give, by interaction with water, an amount 
of oxygen which will just suffice for the oxidation of 10 grams 
of mercury to mercuric oxide, HgO. Ans. 1120 c.c. 




104 


CHEMICAL CALCULATIONS 


171. Calculate the volume of chlorine, at standard condi¬ 

tions, necessary to give an amount of potassium chlorate (by 
interaction with a hot solution of potassium hydroxide) which 
would just suffice, in its decomposition into oxygen and potas¬ 
sium chloride, for the oxidation of 5 grams of hydrogen to 
water. Ans. 55,446 c.c. 

172. 100 grams of iron pyrites, FeS 2 , were roasted to ferric 

oxide, Fe 2 0 3 , and sulphur dioxide. The sulphur dioxide was 
then taken up by sodium peroxide, Na 2 0 2 , to form sodium 
sulphate, and this product treated with a solution of barium 
chloride, BaCl 2 . What weight of barium sulphate was pre¬ 
cipitated? Ans. 389.1 grams. 

173. 20 grams of nitrogen were carried through the follow¬ 
ing series of reactions. Calculate the resulting volume of 
nitrous oxide, N 2 0, at standard conditions. 

3 Mg + N 2 = Mg 3 N 2 
Mg 3 N 2 + 6 H 2 0 = 3 Mg (OH) a + 2 NH 3 
NH 3 + HN0 3 = NH 4 N0 3 
NH 4 N0 3 = N 2 0 + 2 H 2 0 

Ans. 50,237 c.c. 

174. Calculate the relative weights of sodium chlorate and 
potassium chlorate necessary to give the same volume of oxygen. 

Ans. 100 (NaCIO*): 115 (KC10 3 ). 

175. Calculate the relative weights of potassium chlorate 

and perchlorate, KC10 4 , necessary to give the same volume of 
oxygen. Ans. 100 (KC10 3 ): 84.8 (KC10 4 ). 

176. Compare the weights of aluminium and zinc neces¬ 

sary for the production of equal weights of hydrogen by inter¬ 
action with an acid. Ans. 100 (Al); 361.8 (Zn). 

177. Compare the weight of calcium nitride, Ca 3 N 2 (in its 

interaction with water), and the weight of ammonium chloride 
(in its interaction with a base), necessary to give the same 
weight of ammonia. Ans. 100 (Ca 3 N 2 ): 72.2 (NH 4 C1). 

178. What relative weights of cupric oxide, CuO, and 

cuprous oxide, Cu 2 0, are procurable from the same weight of 
copper? Ans. 100 (CuO): 89.95 (Cu 2 0). 


CHEMICAL EQUATIONS 


105 


179. In the interaction of methane, CH 4 , and chlorine, 2 
grams of the former required 17.7 grams of the latter. Calcu¬ 
late the reaction-quantity of chlorine per molecule of methane. 

Ans. 2 Cl 2 . 

180. 2 grams of hydrogen sulphide, H 2 S, decolorized 5.8 

grams of potassium dichromate, K 2 Cr 2 0 7 (in acid solution). 
Calculate the reaction-quantity of the former required per 
molecule of the latter. Ans. 3 H 2 S. 

181. 2.4 grams of ammonia, NH 3 , reduced 17 grams of hot 
cupric oxide, CuO, to copper. Calculate the reaction-quantity 
of cupric oxide required per molecule of ammonia. 

Ans. 1^ CuO. 

Note. — In order to avoid fractional quantities we may here mul¬ 
tiply by two and obtain 3 CuO per 2 NH 3 . 

182. What weight of chrome-alum, K 2 S0 4 . Cr 2 (S0 4 ) 3 . 

24 H 2 0, may be obtained from 20 grams of crystallized potas¬ 
sium sulphate, K 2 S0 4 .6 H 2 0, and an excess of chromium 
sulphate? Ans. 70.76 grams. 

183. What weight of ammonium-magnesium phosphate, 
NH 4 MgP0 4 .6 H 2 0, could be formed from a solution containing 
50 grams of crystallized magnesium sulphate, MgS0 4 .7 H 2 0, 
and an excess of ammonia and sodium phosphate? 

Ans. 49.79 grams. 

184. What weight of iron-ammonium alum, (NH 4 ) 2 S0 4 . 

Fe 2 (S0 4 ) 3 .24 H 2 0, may be formed when 12 grams of ammo¬ 
nium sulphate, (NH 4 ) 2 S0 4 , and 30 grams of ferric sulphate, 
Fe 2 (S0 4 ) 3 , are brought together in concentrated aqueous 
solution? Ans. 72.3 grams. 





CHAPTER X. 


NORMAL SOLUTIONS. 

In reactions between substances in solution no atten¬ 
tion thus far has been given to the actual amount of 
substance contained in a definite volume of solvent. 

In the action between an acid, furnishing hydrogen- 
ion, and a base, furnishing hydroxide-ion, the point 
of neutralization is reached when equal quantities of 
these two kinds of ions — chemically equivalent — are 
present, and the complete removal of both in the form 
of the compound water as a slightly ionized substance 
is effected. The detection of this point in solution is 
readily accomplished through the use of an indicator , or 
some substance which shows a change in color by the 
merest trace of either the one or the other of these two 
ions. The operation of ascertaining this end-point is 
called titration. Naturally it may be applied to the 
determination of other end-points, or points of comple¬ 
tion of definite chemical reactions in solution, as well as 
to this process of neutralization. 

The neutralization of hydrochloric acid by sodium hy¬ 
droxide is shown by the following equation: 

Na* + OH' + H* + C1' =Na* + Cl' + H 2 0 

23 + 17.01 1.01 + 35.46 _ 23 + 35.46 18.02 

4001 36^47 58.46 + 18.02* 

A solution which contains 40.01 parts by w r eight of 
sodium hydroxide will exactly neutralize one which con¬ 
tains 36.47 parts by weight of hydrogen chloride. This 
106 







NORMAL SOLUTIONS 


107 


follows from the fact that in the former there are 17.01 
parts of ionizable hydroxyl and in the latter 1.01 parts of 
ionizable hydrogen, — the exact proportions of these two 
substances necessary for the formation of water. When 
equal volumes of these solutions neutralize each other, 
then the concentration of ionizable hydrogen in the acid 
solution is equal to the concentration of ionizable 
hydroxyl in the solution of the base; i.e., the relative 
amounts of each are directly proportional to 1.01 and 
17.01 respectively. The liter has been adopted as the 
standard volume for reactions in solution; when a gram- 
molecular weight of a substance is contained in this vol¬ 
ume we have what is called a gram-molecular solution or 
more commonly a Molar Solution. Some definite tem¬ 
perature, circa 20°, is usually understood. 

From the reaction between sodium hydroxide and 
hydrochloric acid, and from the definitions just given, we 
are aware that one liter of a molar solution of the former 
will exactly neutralize one liter of a molar solution of 
the latter; consequently any fractional part of the one 
solution will neutralize this same fractional part of the 
second solution. 

In the neutralization of this same base by a dibasic 
acid, such as sulphuric acid, the following equation comes 
into consideration: 

2 Na*+ 2 OH' + 2 H # + SO/=2 Na*+SO/+ 2 H 2 0 
2(23) +2(17.01) +2(1.01) +96.07 = 2(23) +96.07 + 2(18.02) 

From this it is evident that a molar solution of sulphuric 
acid, with 98.09 grams of the acid per liter, will contain 
2.02 grams of ionizable hydrogen, a quantity that re¬ 
quires 34.02 grams (2 X 17.01) of ionizable hydroxyl for 
its complete neutralization. This quantity of hydroxyl 
is furnished, as the equation indicates, through the use 


108 


CHEMICAL CALCULATIONS 


of two gram-molecular weights of sodium hydroxide. If 
we were dealing with molar solutions of these substances, 
two volumes of the sodium hydroxide solution would be 
required for the neutralization of one volume of a molar 
sulphuric acid solution. 

By reason of this variation in the number of ionizable 
hydroxyl and hydrogen groups in the various substances, 
it is found advisable to base our standard solutions upon 
the exact number of these univalent groups which they 
contain, rather than upon the entire molecular weight of 
the substance itself. A solution which contains in one 
liter exactly 1.01 grams of ionizable hydrogen is taken 
as the standard for acids, while that which contains in 
one liter exactly 17.01 grams of ionizable hydroxyl is 
taken as the standard for bases. These two solutions 
are, volume for volume, always equivalent and may be 
termed Equivalent Normal Solutions or, as is more gen¬ 
erally the case, Normal Solutions. 

The molar solution of sodium hydroxide is identical, of 
course, with its normal solution. The molar solution of 
sulphuric acid contains twice as much ionizable hydrogen 
as is required for its normal solution, — a fact indicated 
by the use of two volumes of the molar sodium hydroxide 
solution above to neutralize only one volume of this acid. 
We are therefore required to dissolve one-half of the 
gram-molecular weight (98.09) of sulphuric acid, or 
49.04 grams, in water and bring this to one liter in order 
to obtain its true normal solution. 

In the same manner a base such as barium hydroxide, 
Ba(OH) 2 , with a molecular weight of 171.39, will con¬ 
tain in its molar solution 171.39 grams of substance of 
which 34.02 grams is ionizable hydroxyl. A definite 
volume of this molar solution would neutralize two 
volumes of a normal hydrochloric acid solution; conse- 


NORMAL SOLUTIONS 


109 


quently to obtain a solution of 17.01 grams of hydroxyl 
to the liter (a normal solution) we should need to dis¬ 
solve 171.39/2 grams of the substance in a liter of solu¬ 
tion. A solution of this concentration is here unattain¬ 
able, as the solubility falls below the value required. In 
such cases various degrees of dilution are used, as will be 
indicated below. 

Solutions that furnish neither hydrogen- nor hydrox¬ 
ide-ion are considered normal when they contain, per 
liter, an equivalent of 1.01 grams of hydrogen or 17.01 
grams of hydroxyl. This signifies that in the reactions 
in which they are concerned they will have, per liter, the 
power of combining with or displacing, either directly or 
indirectly, these proportional amounts of hydrogen or 
hydroxyl. Thus, in the action of hydrochloric acid upon 
sodium carbonate: 

Na 2 C0 3 + 2 HC1 = 2 NaCl + H 2 0 + C0 2 , 

it is observed that one gram-molecular weight of the car¬ 
bonate brings into the reaction 2.02 grams of hydrogen; 
consequently one-half of its gram-molecular weight (106), 
or 53 grams of sodium carbonate, is required in 1 liter 
of its normal solution. 

Normality Factors. — For chemical purposes it is not 
necessary to bring every solution to the same standard of 
concentration, — the normal solution. The variations 
from the true normality may be readily expressed by 
fractions, or factors, which designate at once the actual 
concentration of the solutions in terms of the normal. 
Thus a molar solution of sulphuric acid contains twice 
what a normal solution should contain. Its normality, 
therefore, is 2, and is expressed as 2 N. A solution con¬ 
taining 0.365 gram of hydrogen chloride per 100 c.c. would 
contain 3.65 grams per liter. This is 1/10 of what a 


110 


CHEMICAL CALCULATIONS 


normal solution contains; hence the solution is N/10 or 
0.1 N, (deci-normal). 

By another method of procedure, this solution, contain¬ 
ing 0.365 gram of hydrogen chloride per 100 c.c., may be 
compared directly with the amount required in a liter of 
a normal solution, namely, 36.5 grams: —36.5: 0.365 = 
1 : x. The factor ( x) is here 0.01, hence this weight (0.365 
gram) of hydrogen chloride would be contained in 1/100 
of 1000 c.c., or 10 c.c. of the normal acid. As it actually 
occurs in 100 c.c., then the solution in question is 10/100 N 
or N/10; that is, 100 c.c. of this solution is necessarily 
equivalent to 10 c.c. of the normal solution. 

By reason of the equivalence between equal volumes of 
normal solutions we can readily calculate the normality of 
any solution if we are given the normality of that solution 
which is to be titrated against it. 

Example 48 . —100 c.c. of N/2 sodium hydroxide solution 
were required in the neutralization of 400 c.c. of an unknown 
acid solution. Calculate the normality of this acid? 

Here, of course, 100 c.c. of N/2 solution is the equivalent 
of 50 c.c. of a normal solution; i.e., in 100 c.c. of N/2 
sodium hydroxide solution we have 100/1000 or 1/10 of 
17.01/2 grams, or 17.01/20 gram, of hydroxyl, which is 
exactly the amount contained in 1/20 of a liter (50 c.c.) 
of a normal solution containing 17.01 grams per liter. In 
order to find the normality of the unknown solution it will 
be necessary to get some expression for it in terms of the 
known or normal solution. For example, 400 c.c. of this 
unknown solution neutralize 50 c.c. of the normal. By 
reduction, 1 c.c. neutralizes 1/8 c.c. of the normal. The 
relative volumes are equivalent; hence 1 liter of the 
unknown solution must neutralize, and possess an equiva¬ 
lent amount of substance to, that contained in one-eighth 


NORMAL SOLUTIONS 


111 


of a liter of normal sodium hydroxide solution. It is, 
therefore, but 1/8 N, a fact indicated at once by the 
factor obtained for the equivalent of that unit value 1 c.c. 
above. If we had taken as the unit value 1 c.c. of the 
known solution, we should have obtained 8 as a factor 
denoting the number of cubic centimeters of the unknown 
acid solution equivalent in value to 1 c.c. of the known. 
This comes to the same end and indicates the strength of 
1 c.c. of the unknown solution as 1/8 that of the normal. 

Calculation of Normality by Simple Proportion. — A 
very simple method for calculations of this sort rests upon 
the consideration of the proportionality which exists 
between the normality factors and the volumes for these 
equivalent solutions. 

Example Jfi. — 400 c.c. of N/8 acid solution neutralized 100 
c.c. of an unknown alkaline solution. Calculate the normality 
of the alkali. 

Now the volume of a solution when multiplied by its 
normality factor gives, as we have seen, the equivalent 
volume in terms of its normal solution. As all of these 
solutions are balanced or titrated to an end-point which 
signifies that equivalent quantities of the various sub¬ 
stances are present, we may at once place the two expres¬ 
sions for the two solutions as equal to each other. Thus: 

400 X 1/8 = 100 X x. 

All of this is in exact accord with our premises which make 
it necessary for equal volumes to neutralize equal volumes 
when an equivalent amount of substance is present in each. 
By separating these terms of the equation into means and 
extremes of a simple proportion (for example by dividing 
through by the quantity 100 X 1/8) we obtain 
400 x 


112 


CHEMICAL CALCULATIONS 


This is synonymous with saying that the normality factors 
of the two equivalent solutions stand to each other in an 
inverse proportion to the corresponding volumes required 
in the titration: 

100 : 400 = 1/8 : z. 

By calculation x is found to be 1/2, i.e., the alkaline solu¬ 
tion is 0.5 normal. 

Calculation of the Weights of Substances Present in 
Standard Solutions. — When the normality of an unknown 
solution has been determined it is often desirable to find 
the exact amount of substance in a given volume of this 
solution. For example, in the last paragraph the normal¬ 
ity of the alkaline solution was found to be N/2. If it is 
now desired to learn the amount of sodium hydroxide 
actually present in the 100 c.c. of solution, we need only 
take the proportional amount of sodium hydroxide in 
a liter of N/2 solution (20 grams) as is indicated by the 
fractional part which this volume is of 1000, —100/1000 
or 1/10, i.e., 1/10 of 20 grams, or 2 grams. By simple 
proportion a comparison of the volume relations with the 
corresponding weights of the substances gives the follow¬ 
ing: 

1000 : 100 = 20 : z. 

Standardization of Solutions by Gravimetric Means. — 

For our standard solutions, it is usually customary to 
dissolve a certain calculated amount of substance in water 
and bring the volume up to the desired mark by the 
gradual addition of more water. As these solutions may 
vary slightly from the true values, it is always desirable 
to standardize them by purely chemical means, such as 
titration against certain accurately prepared solutions, or 
if possible by the formation of precipitates in a known 
volume of their solution. These precipitates when dried 


NORMAL SOLUTIONS 1J.3 

and weighed serve as a means for the calculation of the 
exact normality. 

Example 50. — 100 c.c. of a hydrochloric acid solution, made 
up approximately to 1/5 N, gave, when treated with a slight 
excess of a silver nitrate solution, 3.22 grams of silver chloride. 
What is the normality of the acid? 

From the equation: 

AgNO s + H£I = AgCl + HN0 3 , 
the proportionality between all amounts of hydrogen 
chloride and silver chloride are indicated by the ratio: 
AgCl : HC1 
(107.88 + 35.46) : 36.47. 

From this ratio the weight of hydrogen chloride, 0.8195 
gram, corresponding to the weight of silver chloride, 3.22 
grams, is easily calculated: 

AgCl : HC1 = 3.22 : x- 
143.34 : 36.47 = 3.22 : 0 .8195,. 

This weight of hydrogen chloride is found present in 100 c.c. 
of solution. In one liter we shall have 8.195 grams, whereas 
we should have 36.47 grams if it were a normal solution. 
The fraction 8.195/36.47 represents then the normality, 
expressed decimally as 0.22^7 N, and gives to the solution 
a value somewhat higher tnan that estimated, (N/5). 

Standardization of Solutions by Volumetric Means. — In 
place of the method of precipitates another very instruct¬ 
ive method is applicable in standardization; chiefly with 
acids. This consists in measuring the volume of a gas 
evolved by the action of some substance upon a known 
volume of the acid solution. 

Example 51. —250 c.c. of an acid solution gave, when acted 
upon by zinc, 560 c.c. of hydrogen (calculated at standard con¬ 
ditions of temperature and pressure). What is the normality 
of the acid. 


114 


CHEMICAL CALCULATIONS 


560 c.c. of hydrogen from 250 c.c. of solution would 
mean, of course, 2240 c.c. from one liter of the solution. 
By definition a normal solution of an acid is one that 
contains 1.01 grams of ionizable hydrogen per liter. The 
gram-molecular weight (2.02) of hydrogen occupies a 
volume of 22,400 c.c. at standard conditions. The 
volume occupied by 1.01 grams therefore will be just 
one-half this, or 11,200 c.c. Accordingly every liter of a 
normal acid solution must contain that weight of hydro¬ 
gen which when set free will occupy 11,200 c.c. under the 
standard conditions of temperature and pressure. 

It is only a simple step to calculate the volume of 
hydrogen per liter when we have given the volume for 
any fraction of a liter. In the problem above, 250 c.c. 
of solution evolved 560 c.c. of hydrogen; consequently 
1 liter will evolve 2240 c.c. of this gas: 250 :1000 = 560 : x. 
As a liter of normal acid should give 11,200 c.c., the 
solution in question is less than normal, and in accord¬ 
ance with the ratio 

11,200 : 2240 = 1 : x } or x = 1/5, 

i.e.j the acid is 0.2 N. In other words, the normality is 
expressed by the fraction, or factor, which the volume of 
hydrogen evolved per liter makes with the total volume 
of hydrogen that can be evolved from a liter of the normal 
acid. 

If the hydrogen is measured at room temperature and 
pressure, it is only necessary to calculate the volume it 
would occupy at the standard conditions and thus make 
possible the comparison between this and the standard 
volume per liter. If this is not done, then the standard 
volume per liter (11,200 c.c.) must be calculated to the 
conditions of the experiment under which the gas is 
measured. 


NORMAL SOLUTIONS 


115 


Similarly, the evolution of other gases by chemical 
action, from known volumes of solutions, may serve for 
the estimation of the normalities of these solutions. For 
example, a solution of sodium carbonate may be treated 
with an acid and the carbon dioxide set free measured. 
The following equation is here involved: 

Na 2 C0 3 + H 2 S0 4 = Na 2 S0 4 + C0 2 + H 2 0. 

The direct proportionality between the reaction-quan¬ 
tities concerned is expressed by the ratio C0 2 /Na 2 C0 3 
or 44/106. This shows that for every gram-molecular 
weight of carbon dioxide (44) we must estimate the pres¬ 
ence of one gram-molecular weight of sodium carbon¬ 
ate (106). Now a normal solution of sodium carbonate 
contains only one-half of the gram-molecular weight in 
1 liter; a fact readily determined by its titration with a 
normal solution of any acid. This is indicated in the 
equation above, wherein we note that 2.02 grams of 
ionizable hydrogen (2 H* + SO/') are required for the 
complete action, and consequently only one-half of the 
gram-molecular weight of the carbonate, 53 grams, can 
be equivalent to 1.01 grams of hydrogen. The amount 
of carbon dioxide evolved from one gram-molecular 
weight of the carbonate is one gram-molecular weight, 
or a volume of 22,400 c.c.; hence from a normal solu¬ 
tion with one-half the gram-molecular weight of car¬ 
bonate, we should have, per liter, just 11,200 c.c. of 
this gas. 

When the volume of carbon dioxide evolved from a 
definite volume of solution is known we need only to 
calculate the volume of gas evolved, per liter, and com¬ 
pare this volume at standard conditions with the stand¬ 
ard volume, 11,200 c.c. The ratio to this value gives the 
ratio to unit normality. 


116 


CHEMICAL CALCULATIONS 


Example 52. — What is the normality of a sodium carbonate 
solution, 125 c.c. of which evolved 350 c.c. of carbon dioxide 
(at standard conditions) when treated with an excess of acid? 

Since 125 c.c. of the solution gave 350 c.c. of the gas, 
1000 c.c. will give 2800 c.c. as determined from the pro¬ 
portion 125 : 1000 = 350 : x. A normal solution should 
evolve 11,200 c.c. of this gas; consequently the solution 
in question is only 2800/11,200 or 1/4 N. 

The evolution of carbon dioxide in the equation above 
may serve equally well in determining the normality of 
the sulphuric acid used. Each liter of normal sulphuric 
acid will liberate 11,200 c.c. of the gas. The ratio of 
comparison, therefore, is carried out just as described in 
the previous paragraph. 

Comparison of Solutions with Standard. — When a 

solution is once standardized other solutions may be 
standardized by comparison with it either directly or 
indirectly. In the case of a second acid solution it is 
necessary to ascertain what volumes of both this acid 
and our standard acid are required for the neutralization 
of equal volumes of some alkaline solution. These two 
equivalent volumes are then compared just as in the 
preceding examples. 

Example 53. — 10 c.c. of 1.5 N hydrochloric acid neutralized 
40 c.c. of an alkaline solution. 50 c.c. of an unknown sulphuric 
acid solution neutralized 80 c.c. of this same alkaline solution. 
Calculate the normality of the sulphuric acid. 

50 c.c. of the sulphuric acid neutralized 80 c.c. of the 
alkali; 25 c.c. of the acid would neutralize 40 c.c. of the 
alkali. This is the same volume of alkali neutralized by 
10 c.c. of 1.5 N hydrochloric acid; hence these two 
volumes of the acid solutions must be equivalent: 

10 c.c. X 1.5 N HC1 = 25 c.c. X (x) N H 2 S0 4 . 


NORMAL SOLUTIONS 


117 


By proportion: 

25 : 10 = 1.5 :*, or * = 3/5. 

Hence the normality of the sulphuric acid is found to be 
3/5, i.e., 0.6 N. 

Adjustment of Solutions to a Desired Standard. — 

When a solution has been standardized and found to 
vary somewhat from the estimated normality, it is cus¬ 
tomary to calculate the amount of substance (if too 
dilute) or water (if too concentrated) that will bring the 
solution to the desired normality. The latter, which con¬ 
stitutes the more simple case, is illustrated in the follow¬ 
ing example: 

Example 54. — A solution of hydrochloric acid was found to 
have a normality of 1.05. What volume of water must be 
added to 400 c.c. of this solution to make it exactly normal? 

1 c.c. of 1.05 N hydrochloric acid is equivalent by 
definition to 1.05 c.c. of a normal hydrochloric acid 
solution. Therefore, to make this more concentrated 
acid normal, we need only add water until 1.05 c.c. of 
the diluted solution will be exactly equivalent to 1.05 c.c. 
of a normal solution; in other words, 

1.05 c.c. — 1 c.c. = 0.05 c.c., 

or that volume of water required per cubic centimeter of 
the 1.05 N acid. 400 c.c. will require 400 X 0.05 c.c., or 
20 c.c. Therefore, when 400 c.c. of this 1.05 N acid are 
diluted with 20 c.c. of water, the final 420 c.c. will be just 
normal. 

By comparison of the actual weights of hydrogen chlo¬ 
ride in equal volumes of these acids and the direct pro¬ 
portionality between these weights and the corresponding 
volumes which are equivalent, volume for volume, we may 
calculate the amount of dilution necessary to bring any 


118 


CHEMICAL CALCULATIONS 


known solution to a desired normality. Thus in 1 c.c. of 
normal hydrochloric acid there is 0.0365 gram of hydro¬ 
gen chloride, and in 1 c.c. of 1.05 N hydrochloric acid 
there is 1.05 X 0.0365 gram of hydrogen chloride. There¬ 
fore 

0.0365 : (1.05 X 0.0365) = 400 : x, or * = 420. 

This gives the volume to which 400 c.c. of 1.05 N acid 
must be diluted. 

When a solution is found too dilute, a similar calcula¬ 
tion will give the amount of water that should be removed 
from the volume in question. Without resorting to this 
procedure, it is found better to add to the entire volume 
that weight of substance (usually a well-defined salt) 
necessary to make with this excess of water a solution of 
the desired normality. Any change in volume due to 
process of solution of the salt may be neglected. 

The calculation of results through reactions which in¬ 
volve solutions of definite concentration follows the general 
outlines presented in Chapter IX. The amount of any 
substance present in a required volume of a standard 
solution is to be considered, of course, with reference to 
its corresponding reaction-quantity. 

The method of determining the amount of a substance 
necessary to complete a given reaction with a known 
amount of some other substance contained in a definite 
volume of solution is called “ Volumetric Analysis.” When 
no reference is made to volume relations, but calculations 
are made upon the weighed quantities which come under 
consideration, we have the more common “ Gravimetric 
Analysis.” These two form the basis of work in quanti¬ 
tative chemical analysis. 


NORMAL SOLUTIONS 


119 


PROBLEMS. 

185 . 400 c.c. of N/4 potassium hydroxide solution were 
required for the neutralization of 600 c.c. of an unknown acid 
solution. Calculate the normality of this acid solution. 

Ans. N/6. 

186 . 500 c.c. of N/10 acid solution were required for the 
neutralization of 25 c.c. of a solution of sodium hydroxide. 
Calculate the normality of this latter solution. Ans. 2 N. 

187 . 220 c.c. of N/20 acid solution were required for titra¬ 
tion with 124 c.c. of a solution of barium hydroxide. Calcu¬ 
late the normality factor of the barium hydroxide solution. 

Ans. .0887 N. 

188. What volume of N/10 acid solution will be required in 
the titration of 440 c.c. of N/4 sodium hydroxide solution? 

Ans. 1100 c.c. 

Note: — The method by simple proportion, with one of the 
volumes as the unknown term, will be found to serve well in such 
examples. 

189 . What volume of N/6 alkaline solution will be required 
in the titration of 254 c.c. of N/10 acid solution? 

Ans. 152.4 c.c. 

190 . Calculate the weight of hydrogen chloride present in 
400 c.c. of a hydrochloric acid solution which required 320 c.c. 
of N/4 alkaline solution for titration. Ans. 2.918 grams. 

191 . Calculate the weight of sulphuric acid present in 

150 c.c. of a solution which required 48.1 c.c. of 0.78 N alkali 
for titration. Ans. 1.84 grams. 

192 . An excess of silver nitrate solution was added to 350 

c.c. of a solution of hydrochloric acid. The precipitate of silver 
chloride weighed 7.54 grams. Calculate the normality of the 
acid. Ans. 0.1503 N. 

193. A slight excess of barium chloride solution was added 

to 400 c.c. of a solution of sulphuric acid. From the weight of 
barium sulphate, BaS0 4 , precipitated, 4.12 grams, calculate 
the normality of the acid. Ans. 0.0882 N. 


120 


CHEMICAL CALCULATIONS 


194. 600 c.c. of a sulphuric acid solution, when acted upon 
by an excess of zinc, evolved 1242 c.c. of hydrogen (at standard 
conditions). Calculate the normality of the acid. 

Ans. 0.1848 N. 

195. 440 c.c. of an acid solution, when acted upon by an 

excess of zinc, evolved 2430 c.c. of hydrogen, measured over 
water at 21° and 747.5 mm. pressure. Calculate the nor¬ 
mality of the acid. Ans. 0.4408 N. 

196. Calculate the normality of an acid solution, 600 c.c. 

of which, when acted upon by an excess of sodium carbonate, 
evolved 2100 c.c. of carbon dioxide (calculated to standard 
conditions). Ans. 0.3125 N. 

197. Calculate the normality of a solution of potassium 

carbonate, 200 c.c. of which, when treated with an excess of 
acid, evolved 4502 c.c. of carbon dioxide (calculated to 
standard conditions). Ans. 2.01 N. 

198. An excess of iron sulphide, FeS, was added to 500 c.c. 

of a solution of sulphuric acid. The volume of hydrogen sul¬ 
phide set free measured 4640 c.c. (at standard conditions). 
Calculate the normality of the acid. Ans. 0.8285 N. 

199. An excess of sodium sulphite, Na 2 S0 3 , was added to 
400 c.c. of a solution of hydrochloric acid. The volume of 
sulphur dioxide set free measured 5600 c.c. (at standard con¬ 
ditions). Calculate the normality of the acid. Ans. 1.25 N. 

200. 1400 c.c. of ammonia (calculated to standard condi¬ 

tions) were passed into 500 c.c. of N/2 hydrochloric acid solu¬ 
tion. Calculate the normality of the hydrochloric acid still 
present. Ans. N/4. 

201. 210 c.c. of carbon dioxide (at standard conditions) 

were passed into 250 c.c. of N/10 barium hydroxide solution. 
Calculate the normality of the barium hydroxide solution still 
present. Ans. N/40. 

202. 50 c.c. of N/5 hydrochloric acid solution neutralized 
40 c.c. of an unknown alkaline solution. 300 c.c. of a sul¬ 
phuric acid solution neutralized 60 c.c. of this same alkaline 
solution. Calculate the normality of the sulphuric acid. 

Ans. 0.05 N. 


NORMAL SOLUTIONS 


121 


203 . 200 c.c. of a barium hydroxide solution were required 

in the titration of 40 c.c. of an acid solution. 100 c.c. of this 
acid solution exactly neutralized 80 c.c. of N/2 alkaline solu¬ 
tion. Calculate the normality of the barium hydroxide solu¬ 
tion. Ans. 0.08 N. 

204. A solution of hydrochloric acid is desired to be made 
exactly normal. 40 c.c. of the solution neutralized 50 c.c. 
of 0.84 N sodium hydroxide solution. Calculate the volume of 
water that must be added per 100 c.c. of the acid solution. 

Ans. 5 c.c. 

205. A solution of sodium hydroxide is desired to be made 

exactly 0.5 N. 32 c.c. of the solution at hand were required 
for the titration of 28 c.c. of 0.8 N hydrochloric acid. Calcu¬ 
late the volume of water that must be added per 100 c.c. of the 
alkaline solution. Ans. 40 c.c. 

206. A solution of sodium carbonate is desired to be made 

exactly 0.05 N. 24 c.c. of the solution at hand neutralized 
9.6 c.c. of 0.12 N hydrochloric acid solution. Calculate the 
weight of anhydrous salt, Na 2 C0 3 , that must be added per 100 
c.c. of solution. Ans. 0.0106 gram. 

207 . What weight of iron will be required for interaction 
with 400 c.c. of N/5 hydrochloric acid? Ans. 2.234 grams. 

208 . What weight of sodium carbonate will be required for 
interaction with 600 c.c. of N/8 sulphuric acid? What volume 
of carbon dioxide (at standard conditions) will be evolved? 

Ans. 3.975 grams. 

840 c.c. C0 2 . 

209 . What weight of sodium hydrogen carbonate, NaHCO s , 

will be required for interaction with 600 c.c. of N/8 sulphuric 
acid? What volume of carbon dioxide (at standard conditions) 
will be evolved? Ans. 6.3 grams. 

840 c.c. CO a . 

210. Calculate the weight of crystallized oxalic acid, 

C 2 H 2 0 4 .2 H 2 0, required for a solution which is to be made up 
to 500 c.c. in volume at N/2. Ans. 31.51 grams. 

211. Calculate the volume of nitric oxide, NO (at standard 

conditions) that could be evolved by the action of copper upon 
1000 c.c, of a 7 N nitric acid solution. Ans. 39,200 c.c. 


CHAPTER XI. 


COMBINATIONS BETWEEN GASES BY VOLUME. 

When expressed in grams the molecular weights of all 
substances in the state of vapor occupy a volume of 
22,400 c.c. at the standard conditions of temperature and 
pressure. In Chapter IX it was observed that a molecular 
equation offered for this reason an insight into the volume 
relations of the various substances concerned in a given 
reaction. 

Molecular Volumes. — These volumes comply naturally 
with the laws relating to gases, and, further, are subject to 
the operation of those properties, characteristic of each 
substance, which may here be brought into consideration 
through the conditions of the experiment. Thus the con¬ 
densation of a gas or vapor to the liquid or solid state, or 
its solution in, or combination with, various substances 
which may be present, will remove it completely from 
further considerations. 

As a simple illustration of these facts an excellent ex¬ 
ample is found in the combination of hydrogen with 
oxygen: 

2 H 2 + 0 2 = 2 H 2 0. 

By molecules: 2 +1 =2. 

By gram-molecular 
volumes: 

2 (22,400) c.c. + 1 (22,400)c.c. = 2 (22,400) c.c. + 

(22,400 c.c. contraction). 


122 


COMBINATIONS BETWEEN GASES BY VOLUME 123 

Reduced through the 
common term, 22,400: 

2 vol. + 1 vol. = 2 vol. + (1 vol. contr.). 

Each gram-molecular weight may be represented by 
one volume, i.e., its gram-molecular volume. The coeffi¬ 
cients or integers which designate the number of gram- 
molecular weights will stand likewise for the number of 
gram-molecular volumes involved in the reaction. The 
combination of gases by volume expressed thus by simple 
numbers has been developed experimentally and is com¬ 
prehended in Gay-Lussac’s well-known Law of Combining 
Volumes. 

In the reaction above we observe that 2 volumes of 
hydrogen and 1 volume of oxygen unite to form 2 volumes 
of aqueous vapor, the temperature of 100° or above, 
and the observed pressure, remaining constant through¬ 
out. This diminution in the total volume of the gaseous 
components when transformed into aqueous vapor 
has been considered in the study of Avogadro’s hy¬ 
pothesis. 

If the temperature falls below 100° a further contraction 
in volume occurs. Since this is due to the condensation 
of aqueous vapor to the liquid state, and increases accord¬ 
ingly with a lowering of temperature, there is left, eventu¬ 
ally, no volume of gaseous product. The water formed 
becomes associated with the liquid over which the gases 
are measured, and consequently drops out of further 
consideration through the equalization or adjustment of 
the levels within and without the vessel to bring all to 
uniform pressure. Of course a small amount of water 
remains in the vapor state, even at a low temperature; 
corrections for this are made by reference to a table of 
aqueous vapor tensions (Appendix II). 


124 


CHEMICAL CALCULATIONS 


The volume of the dry gaseous product, which in this 
case is theoretically nil, may be expressed in this manner: 
2 H 2 + 0 2 = 2 H 2 0. 

Molecular volumes: 2 +1 = 2 + (1 vol. contr.). 

Molecular volumes 

below 100°: 2 +1 — 0 + (3 vol. contr.). 

The Relation of Molecular Volume to a Definite Volume- 
Unit. — In the study of these volume-changes in known 
reactions it is always necessary to determine the exact 
relation which any measured volume of vapor, under 
consideration, bears to the corresponding molecular vol¬ 
ume representing it in the molecular equation governing 
the reaction. 

The coefficients of the quantities in a molecular equation 
determine the ratios between the volumes of these sub¬ 
stances when in the state of vapor. The coefficient unity 
stands, of course, for the unit of molecular volume (22,400 
c.c.), a volume corresponding to a gram-molecular weight 
of substance. Any fractional part of the gram-molecular 
weight of a substance will occupy, therefore, under stand¬ 
ard conditions, a volume denoted by this same fractional 
part of the gram-molecular volume, 22,400 c.c. What¬ 
ever may be this fractional part of the molecular quantity 
of a substance under consideration in a reaction, all other 
substances possible of interaction with this one must be 
correspondingly reduced from their own molecular quan¬ 
tities. 

These coefficients, therefore, are constant /or any known 
molecular equation and stand for the ratio between the 
volumes of the substances in state of vapor, be they 
gram-molecular volumes or any definite fractional parts 
of these. That volume which is represented by the co¬ 
efficient unity in a molecular equation is indeed the unit 


COMBINATIONS BETWEEN GASES BY VOLUME 125 

upon which all of the other substances, in the state of 
vapor, enter into the reaction. It may be regarded as the 
Volume-Unit for the reaction given. If the coefficient 
representing a certain substance in a molecular equation 
is greater than unity, then, of course, from the actual 
volume of vapor of the substance here concerned, the 
true volume-unit for the equation can be derived only 
by dividing this known volume by the corresponding 
coefficient. 

Calculation of the Volume-Unit from a Single Known 
Volume. — When once the volume-unit for a reaction is 
established, the actual volumes of the several substances 
here entering into combination are easily determined 
from the product of the several coefficients in the molecu¬ 
lar equation by the volume-unit. 

Example 55. — What volume of oxygen will be required to 
burn 300 c.c. of hydrogen and what volume of aqueous vapor 
will result, the temperature of 100° and the atmospheric 
pressure remaining constant throughout? 

2 H 2 + 0 2 = 2 H 2 0. 

Molecular volumes: 

2 +1 = 2 + (1 vol. contr.). 

By placing the molecular volume coefficients out by 
themselves, the relative volumes of the several sub¬ 
stances are indicated. Since we can make our calcula¬ 
tions only from the volumes actually given as data, we 
refer the volume of hydrogen, 300 c.c., directly to the 
coefficient 2. From this we derive the volume 150 c.c. 
(300 -T- 2) as the value for the coefficient of unity, i.e., 
the volume-unit of the reaction. There remains now 
only to substitute this value for each coefficient through¬ 
out the entire molecular equation. 


126 


CHEMICAL CALCULATIONS 


2 H 2 + 0 2 = 2 H 2 0. 

Molecular volumes: 2+1 = 2 + (1 vol. contr.). 
Volumes upon the unit 150 c.c.: 

300 c.c. + 150 c.c. = 300 c.c + (150 c.c. contr.). 

The solution of the problem is clearly seen. 150 c.c. of 
oxygen are required for the combustion of the 300 c.c. of 
hydrogen, while the volume of aqueous vapor that results 
will measure 300 c.c. 

Calculation of Volume-Unit from Known Mixtures of 

Gases. — Since the coefficients or integers represent the 
ratios between the several volumes which enter into 
chemical combination, it must be understood that these 
combinations take place only in accordance with these 
ratios, and that, if any substance is present in amount to 
exceed the stipulated volume, this excess must remain 
unaffected in the reaction; if present in amount less 
than the stipulated volume, the entire reaction must run 
upon a somewhat smaller scale, or volume-unit, for this 
particular substance; with the result that the other sub¬ 
stance or substances concerned will be in excess of the 
corresponding volumes stipulated by the ratios, and 
consequently a fraction of each of these will remain 
unaffected. 

Example 56. — A mixture of 250 c.c. of hydrogen and 150 c.c. 
of oxygen was submitted to the action of an electric spark. 
What was the volume of the product after the explosion, the 
temperature 100° and atmospheric pressure constant? 

2 H 2 + 0 2 = 2 H 2 0. 

Molecular volumes: 2 +1 = 2 + (1 vol. contr.). 

In making hydrogen the basis for our calculations, we 
derive from the volume 250 c.c. the value 125 c.c. as the 
volume-unit. Substituting this value for unity in the 
equation above we obtain the following: 


COMBINATIONS BETWEEN GASES BY VOLUME 127 


Volume f250 c.c. + 125 c.c. = 250 c.c. + (125 c.c. contr.). 
relations,} 


Volumes 

given, 


250 c.c. + 150 c.c. 


Excess in 
oxygen, 


25 c.c. 


The volume of aqueous vapor formed is 250 c.c. The 
volume of oxygen, however, is seen to be in excess by 
25 c.c. of the volume actually required in the reaction; 
hence the apparent volume of product will be increased 
by this same amount: 

250 c.c. + 25 c.c. = 275 c.c., 


which is the volume of gaseous mixture after the explo¬ 
sion. 

In place of hydrogen as the basis for this calculation 
of the volume-unit, we may now choose oxygen. The 
volume given, 150 c.c., will then correspond to a volume- 
unit in the equation; hence by substituting this value 
throughout we obtain 

2 H 2 + 0 2 =2 H 2 0 

2+1 = 2 . 


Volume relations: 300 c.c. + 150 c.c. = 300 c.c. 

Volumes given: 250 c.c. + 150 c.c. 

Here the volume of hydrogen demanded by the equa¬ 
tion for combination with 150 c.c. of oxygen actually 
exceeds the amount of hydrogen at our command. Our 
volume-unit, therefore, has been placed too high. It is 
necessary, then, to reduce these high values until the 
calculated volumes of all the separate components fall 
equal to or under the volumes of substances actually 
present. Such, of course, was true in the selection of 
hydrogen, as above, for the basis of the calculation. 





128 


CHEMICAL CALCULATIONS 


Calculation of True Volume-Unit for a Reaction. — In 

general, we determine a volume-unit for a reaction from 
each of the known volumes of substances present; this 
of course by dividing each volume by the corresponding 
coefficient which represents it in the molecular equation. 
Upon comparison of the volume-units thus derived, the 
smallest will stand for the true unit. In the example 
above the volume-unit upon the hydrogen basis is 125 c.c., 
upon the oxygen basis it is 150 c.c; consequently the value 
125 c.c. alone fulfills the requirements of the reaction. 

Example 57. —100 c.c. of ammonia and 90 c.c. of oxygen 
were exploded. What was the final volume of the product? 
A temperature of 100° and atmospheric pressure constant. 

The reaction proceeds in accordance with the equation: 


4 NH 3 + 3 0 2 = 2 N 2 + 6 H 2 0. 
Molecular 
volumes: 

From the volume of ammonia: 


+ 3=2 + 6 — (1 vol. expan.). 


100/4 = 25 c.c. = volume-unit. 
From the volume of oxygen: 

90/3 = 30 c.c. = volume-unit. 


The smaller value, 25 c.c., must be, therefore, the 
volume-unit for this reaction, while the oxygen will be 
found slightly in excess of the required amount. 

Substituting the value 25 c.c. for unity throughout, we 
obtain 


Volume I 
relations: [ 
Volumes f 
given: j 

Excess: 


4 NH 3 + 3 0 2 = 2 N 2 + 6 H 2 0 

4+3 =2 + 6 —(1 vol. expan.). 

100 c.c. + 75 c.c. = 50 c.c.+ 150 c.c. 


100 c.c. + 90 c.c. 


15 c.c. oxygen. 




COMBINATIONS BETWEEN GASES BY VOLUME 129 

Therefore the volume of product will be equal to 
50 c.c. + 150 c.c., or 200 c.c., plus the excess of oxygen, 
15 c.c., remaining unacted upon, or 215 c.c. in all. 

The determination of this volume-unit in combinations 
between gases is at the basis of all these calculations. In 
examples where the volume of only one substance is given 
it is often difficult to ascertain the volumes of other sub¬ 
stances that may enter into the specified reaction. 

Calculation of Volume-Unit from Measured Volume- 
Changes. — Where calculations are based alone upon the 
volumes of gases that enter into combination, the exact 
proportion of each and every component of the mixture 
must be known; otherwise the volume-units for the several 
reactions cannot be determined. If one or more of these 
factors are unknown, then a study of the various con¬ 
tractions and expansions in the total volume of product 
over that of the original mixture offers a direct method 
for the solution. Heretofore we have concerned ourselves 
only with the components that entered into a reaction. 
The study of the products, however, offers far greater 
possibilities for the reason that the numerous condensa¬ 
tions and absorptions serve to estimate the volumes of 
the many substances possible of formation. Through 
these observations, and a study of the equations involved, 
we may deduce all of the relations between the various 
components that entered into a reaction. 

These conclusions are made possible by the conditions 
of a molecular equation, wherein only molecular volumes 
and multiples of these are concerned. An expansion or 
contraction in total volume of product over the original 
volume must be represented, therefore, by a gain or loss 
in a definite number of molecular volumes. This number 
will be denoted by the difference between the sum of the 
molecular volumes on one side of the equation and the 


130 


CHEMICAL CALCULATIONS 


sum of the molecular volumes on the other. If this 
difference corresponds to a single molecular volume, i.e., 
a volume with coefficient unity, then the expansion or 
contraction observed is in reality the exact measure of 
the volume-unit for the reaction; whereas, if this difference 
is greater or less than unity, the change in volume observed 
must be reduced, through division by the coefficient for 
this difference, when a volume is obtained corresponding 
to a coefficient of unity, i.e., the volume-unit. 

Example 58. —100 c.c. of a sample of air were mixed with 
100 c.c. of hydrogen (an excess) and exploded. After the removal 
of the aqueous vapor by absorption, the volume of dry gaseous 
product read 140 c.c. Calculate the percentage of oxygen in 
this sample of air, the room temperature and pressure remain¬ 
ing a constant throughout. 

2 H 2 + 0 2 = 2 H 2 0. 

2 + 1 = 2+(1 vol. contr.). 

2+1 = 0 +(3 vol. contr.). 

If the conditions were such that all of the water remained 
in state of vapor (100°), the contraction observed, and rep¬ 
resented by a coefficient of unity, would stand for the con¬ 
traction of 1 volume-unit in the equation, and from this 
value the volume of oxygen present — 1 volume-unit — 
would be found equal in value to the contraction itself. 

With the conditions otherwise and the loss of aqueous 
vapor increasing the molecular volume contraction by 2 
molecular volumes (2 volume-units), we note that the 
observed contraction must be due to the loss of 3 volume- 
units from the side of the products. The original mixture, 
100 c.c. + 100 c.c., or 200 c.c., lost 60 c.c. in this reaction 
(200 c.c. — 140 c.c.); hence 60 c.c. represents the 3 volume- 


Molecular volumes 
at 100°: 

Molecular volumes 
at room tempera¬ 
ture: 


COMBINATIONS BETWEEN GASES BY VOLUME 131 

units, or 1 volume-unit will be equal to 20 c.c. From the 
equation, the oxygen consumed is exactly 1 volume-unit 
and hence 20 c.c. This indicates that 20 per cent of the 
sample of air was oxygen. 

Example 59. — A volume of ammonia was mixed with a 
large excess of oxygen and exploded. The expansion in volume 
of product over the original mixture was 50 c.c. What was the 
volume of ammonia? Temperature of 100° and atmospheric 
pressure a constant. 

The equation for this action has been given before: 

4 NH 3 + 3 0 2 = 2 N 2 + 6 H 2 0. 

Molecular j 4+3 = 2+ 6. 

volumes: } 7 vol. = 8 vol. — (1 vol. expan.). 

In this case the expansion is seen to be exactly equal to 
a unit coefficient in volume; consequently 50 c.c. is the 
volume-unit for the reaction. By substituting this value 
in the molecular equation, the volume of ammonia (4 vol¬ 
ume-units), which in its combustion can produce this 
expansion of 50 c.c., is found equal to 4 X 50 c.c. or 
200 c.c. 

From a study of the volume of products, and the alter¬ 
ations in this volume through elimination of certain of 
the substances, we are able to determine the actual com¬ 
position of unknown mixtures of gases. 

Example 60. — 200 c.c. of a mixture of nitrogen and methane 
were exploded with 400 c.c. of oxygen. The volume of the dry 
gaseous product measured 500 c.c. What was the percentage 
of methane in the original mixture? The room temperature and 
pressure a constant. 

CH 4 + 20 2 = C0 2 + 2 H 2 0. 
Molecular volumes: 1 + 2 = 1 + 2 

Molecular volumes, I 

aqueous vapor j 1 +2 =1 + (2 vol. contr.). 

removed: 


132 


CHEMICAL CALCULATIONS 


Here no change in volume results from the explosion when 
the temperature is 100° or over. When the temperature 
falls and the aqueous vapor is condensed we have a new 
relation between the volumes of product and original 
mixture such that a contraction corresponding to 2 volume- 
units occurs. In the example the actual contraction due 
to this removal of the water is 100 c.c., (200 +400 — 500). 
From this we derive the value of a single volume-unit 
as equal to 50 c.c. By reference then to the volume 
equation above, 1 volume-unit, i.e., 50 c.c., represents 
the amount of methane present in the original mixture, 
of which it constitutes 25 per cent (by volume). The 
nitrogen is considered as without action. 

In practice the removal of aqueous vapor is usually not 
carried out. Since at room temperatures the greater part 
of this vapor condenses to the liquid state, the partial 
pressure of this vapor at the observed conditions gives at 
once a means of calculating the volume of dry gaseous 
product. Some definite conditions of temperature and 
pressure are of course taken as a basis. In this manner 
a direct comparison between the volume of product and 
that of the original mixture can be readily made. 

Example 61. — A mixture of 200 c.c. of ethylene and 800 c.c. 
of oxygen (an excess), at 24° and 756.5 mm. pressure, was 
exploded. After the explosion the volume of product read 
649.1 c.c., at 27° and this same pressure. Calculate the per¬ 
centage purity of the ethylene by volume. 

The reaction takes place according to the equation: 

C 2 H 4 + 3 0 2 = 2 C0 2 + 2 H 2 0. 
Molecular volumes: 1+3 = 2+ 2 

Molecular volumes, f 

aqueous vapor \ 1 + 3 = 2 + (2 vol.contr.) 

removed: 


COMBINATIONS BETWEEN GASES BY VOLUME 133 


When allowance is made in this final volume of product 
for the presence of aqueous vapor, which at 27° has a 
tension of 26.5 mm., we have only to calculate the volume 
which the gaseous product will occupy at the original con¬ 
ditions, 24° and 756.5 mm. pressure, the basis thus selected 
for comparison. The expression is 



V' = 620 c.c. 


The original mixture, 200 c.c. + 800 c.c., or 1000 c.c., 
suffered therefore a contraction of 380 c.c. (1000 c.c. — 
620 c.c.). This contraction, due to the removal of the 
aqueous vapor from the side of the products, corresponds 
to 2 volume-units; consequently 1 volume-unit for this 
reaction is equal to 190 c.c., and the volume of ethylene 
concerned (1 volume-unit) is also 190 c.c. We may as¬ 
sume, accordingly, the presence of 10 c.c. of some inert gas 
(e.g. nitrogen) in the original volume (200 c.c.) of ethylene 
taken. This signifies that the sample of ethylene was 
only 95 per cent pure. 

Considerations into which Different Volume-Units may 

Enter. — When a contraction arises from the combined 
effect of two or more reactions it may be impossible to 
derive any one of the volume-units, for, unless the rela¬ 
tive volumes of the several gases going to produce the 
contraction is known, we cannot properly apportion this 
volume of contraction between the several equations. 
In other words, the volume-unit may be different for 
every equation that is brought into consideration. In 
such cases as these it is necessary to bring the final 
products under new conditions, whereby further conden¬ 
sations or absorptions can take place, and the possibility 
of involving separately only one product from one of the 



134 


CHEMICAL CALCULATIONS 


reactions at a time made likely. An individual reaction 
thus concerned yields itself at once to the determination 
of its specific volume-unit. By reference of this volume- 
unit to the several reactions of a problem it is often pos¬ 
sible to determine other volume-units. Still further con¬ 
densations or absorptions through new conditions may be 
necessary, however, to aid in determining the volume-units 
of certain equations when a large number of substances 
are present in the original mixture. When all of these 
values are found the solution of the problem is compara¬ 
tively simple. 

Example 62. — A mixture of nitrogen, hydrogen and carbon 
monoxide, 450 c.c. in volume, was exploded with an excess of 
oxygen, 250 c.c. After the explosion the volume of gaseous 
product measured 500 c.c. With the removal of the aqueous 
vapor the volume of product measured 400 c.c. What was the 
volume of each component in the original mixture? A tem¬ 
perature of 100° and atmospheric pressure considered constant. 


The two equations are as follows: 


2 H 2 

Molecular volumes: 2 
2 CO 

Molecular volumes: 2 


+ 0 2 = 2 H 2 0. 

+ 1 = 2 + (1 vol. contr.). 
+ 0 2 = 2 C0 2 . 

+ 1 = 2 + (1 vol. contr.). 


In this problem the dry gaseous product measured 
400 c.c., i.e., by loss of the aqueous vapor, a contraction of 
100 c.c. was recorded. This contraction is due alone to the 
reaction of hydrogen with oxygen, and stands for the loss 
of 2 volume-units of aqueous vapor in the first volume 
equation. Therefore 1 volume-unit in this equation is 
equivalent to 50 c.c., and 2 volume-units, representing 
the actual amount of hydrogen concerned in the reaction, 
correspond to 100 c.c. 


COMBINATIONS BETWEEN GASES BY VOLUME 135 

In the combustion of hydrogen the contraction actually 
possible corresponds to 1 volume-unit of the reaction. 
This volume-unit has just been determined as 50 c.c. 
The total contraction in the problem, due to the two 
reactions and without elimination of aqueous vapor, is 
recorded as 200 c.c.; hence the difference, 200 c.c. —50 c.c., 
or 150 c.c., represents the contraction due to the reaction 
between carbon monoxide arid oxygen. This contraction, 
as shown in the second volume equation above, is equiva¬ 
lent to 1 volume-unit. Accordingly the volume of the 
carbon monoxide present — 2 volume-units — must have 
been 2 X 150 c.c. or 300 c.c. All told, we derive the 
following composition for the original mixture: 100 c.c. 
of hydrogen, 300 c.c. of carbon monoxide and 50 c.c. of 
nitrogen (considered in these problems as an inert gas). 

The Algebraic Method for the Calculation of Volume- 
Units. — More complicated examples may be given, but 
throughout all the same principles hold. We must first 
determine the volume-unit for each of the reactions under 
consideration before we can calculate the several volumes 
concerned in the problem. When a change in volume of 
product is found to result, not from one single reaction, 
but from a combination of several, and when also, through 
new conditions presented, still further condensations occur, 
which in turn are found to be due to a combination of 
reactions, we find it necessary to give the volume-unit for 
each reaction an algebraic symbol, and solve for the value 
of each unit from the several equations that may be 
constructed. 

Example 63. — A mixture of nitrogen, carbon monoxide, 
methane and ammonia, amounting to 1250 c.c., was exploded 
with an excess of oxygen (1100 c.c.), and the volume of product 
found to measure 2450 c.c. A temperature of 100° and 760 mm. 
constant throughout. After withdrawal of aqueous vapor by 


136 


CHEMICAL CALCULATIONS 


absorption, the volume of the dry gases at the recorded constant 
conditions was 950 c.c. After removal of carbon dioxide (by 
passing gases over lime) the volume read 600 c.c. What was 
the volume of each component in the original mixture? 

The equations for the three reactions are as follows: 

2 CO + 0 2 = 2 C0 2 

Molecular volumes: 2 +1 = 2 + (1 vol. contr.). 

CH 4 + 2 0 2 = C0 2 + 2 H 2 0 
Molecular volumes: 1 +2=1 +2. 

4 NH 3 + 3 0 2 = 2 N 2 + 6 H 2 0 
Molecular volumes: 4 +3 =2 + 6 — (1 vol. expan.). 

An expansion follows the combustion of ammonia 
with oxygen. A contraction due to the combustion of 
carbon monoxide is not sufficient to make up for expan¬ 
sion in the former case, since the final volume, 2450 c.c., 
is 100 c.c. larger than the volume of the original mixture 
of gases. As the contraction and expansion are each 
represented by a volume-unit in their respective reac¬ 
tions, we draw the conclusion that the volume-unit in the 
ammonia reaction is larger by 100 c.c. than the volume- 
unit in the carbon monoxide reaction. The withdrawal 
of aqueous vapor is similarly distributed over two reac¬ 
tions, as is also the withdrawal of carbon dioxide. 

The data from this problem do not yield positive 
information in regard to any single volume-unit of any 
reaction. Combinations between the reactions are of 
course easily intelligible. Thus the withdrawal of the 
water removes 6 volume-units from the ammonia reac¬ 
tion and 2 volume-units from the methane reaction. 
With all the volume-units unknown we may profitably 
study the combination of these units with the idea of 
deriving some one of them and eventually all. For this 


COMBINATIONS BETWEEN GASES BY VOLUME 137 


purpose let x represent the volume-unit of the carbon 
monoxide reaction, y the volume-unit of the methane 
reaction, and z the volume-unit of the ammonia reac¬ 
tion. From the preceding remarks and a study of the 
reactions themselves we may now draw up the following 
equations. 

The expansion is due to the larger value of z compared 
with x: 

(a) z — x = 100. 

The volume of aqueous vapor is represented by 1500 c.c. 
(2450 c.c. — 950 c.c.), and is therefore expressed by 

(6) Gz + 2y = 1500. 


The volume of carbon dioxide corresponds to 2 volume- 
units of x and 1 volume-unit of y and is represented by 
350 c.c. (950 c.c. — 600 c.c.). 

(c) y + 2 x = 350. 

By combining equations ( b ) and (c) to eliminate y, 

(b) 6 0 + 2 y = 1500. 

Twice (c) or 4 x + 2 y = 700, and subtracting 
6 z — 4 x = 800, and subtracting 

four times (a) or 4 z — 4 x = 400 
21 = 400 

2 = 200. 


With the volume-unit of the ammonia equation thus 
derived and equal to 200 c.c., we have only to derive the 
value of x from equation (a) as equal to 100 c.c. and then 
in turn from ( b ) we derive the value of y as equal to 150 c.c. 
Substituting these values in their proper equations the 
volume of carbon monoxide (2 x) is found to be 200 c.c., 
that of methane ( y ) 150 c.c., and that of ammonia (4 z) 
800 c.c., making in all 1150 c.c. Therefore the remaining 




138 


CHEMICAL CALCULATIONS 


volume of the original 1250 c.c., or 100 c.c., is calculated as 
nitrogen. 

Considerations into which the Formation of Non-gase- 
ous Substances Enter. — In the study of those molecular 
equations in which a number of non-gaseous substances 
come under consideration we have simply another case 
of removal of molecular quantities. 

Example 64 . — When phosphorus is burned in a vessel con¬ 
taining 100 c.c. of nitrous oxide, what will be the volume of 
nitrogen left? The room temperature and pressure constant 
throughout. 

The solid phosphorus pentoxide as well as the phos¬ 
phorus fall out of consideration in the volume equation: 

5 N 2 0 + P 2 = 5 N 2 + P 2 0 5 . 

5 N 2 0 =5 N 2 . 

Molecular volumes: 5 =5. 

There is, therefore, no change in volume and 100 c.c. will 
represent also the volume of the nitrogen. 

Again we may consider the reaction of hydrogen 
chloride upon a carbonate: 

Example 65. — An excess of sodium hydrogen carbonate was 
placed in a vessel containing 100 c.c. of hydrogen chloride. 
Calculate the volume of dry carbon dioxide liberated in the 
reaction. Room temperature and pressure a constant. 

NaHCO s + HC1 = NaCl + H 2 0 + C0 2 . 
Molecular volumes: 1 = 1+1. 

We have in this volume equation 1 volume of hydrogen 
chloride liberating, from 1 molecule of sodium hydrogen 
carbonate, 1 volume of carbon dioxide and 1 volume of 
aqueous vapor. In the dry state, therefore, the volume 


COMBINATIONS BETWEEN GASES BY VOLUME 139 

relations of carbon dioxide and hydrogen chloride are as 
1:1. In the case of the normal sodium carbonate the 
volume relations of these dry gases are as 1 : 2. 

Na 2 C0 3 + 2 HC1 = 2 NaCl + H 2 0 + C0 2 . 

2 = 1 . 

The proportional values of carbon dioxide and hydrogen 
chloride are different in the two cases. This, however, is 
dependent upon the nature of the substances concerned 
and is easily explainable by a study of the two reactions. 
Whatever be the complexity of the examples here pre¬ 
sented, they nevertheless may be made to conform to 
very simple interpretations when once the molecular 
equations for the reactions are constructed and a study 
of the volume relations undertaken. 

PROBLEMS. 

212. In the combustion of 400 c.c. of hydrogen with oxygen, 
what volume of oxygen will be required, and what volume of 
aqueous vapor will result? A temperature of 100° and a pres¬ 
sure of 760 mm. constant throughout. Ans. 200 c.c. oxygen. 

400 c.c. vapor. 

213 . A mixture of 300 c.c. of hydrogen and 200 c.c. of oxygen 

was exploded by electric spark. What was the volume of 
product? A temperature of 100° and pressure of 760 mm. 
constant throughout. Ans. 350 c.c. 

214 . A mixture of 300 c.c. of hydrogen and 130 c.c. of oxy¬ 
gen was exploded. What was the volume of product? Tem¬ 
perature of 100° and pressure of 760 mm. constant throughout. 

Ans. 300 c.c. 

215 . A mixture of 420 c.c. of hydrogen and 180 c.c. of 

oxygen was exploded. What was the volume of product after 
the removal of aqueous vapor (by absorption with phosphorus 
pentoxide)? Temperature of 100° and pressure of 760 mm. a 
constant. Ans. 60 c.c. 


140 


CHEMICAL CALCULATIONS 


216. A mixture of 300 c.c. of methane and 150 c.c. of oxy¬ 
gen was exploded. What was the volume of the product after 
the removal of aqueous vapor (by absorption) ? Temperature 
of 100° and pressure of 760 mm. constant throughout. 

Ans. 300 c.c. 

217. A mixture of 250 c.c. of carbon monoxide and 120 c.c. 
of oxygen was exploded. Calculate the volume of gaseous 
product. Temperature and pressure a constant. 

Ans. 250 c.c. 

218. A mixture of 320 c.c. of carbon monoxide and 180 c.c. 
of oxygen was exploded. Calculate the volume of gaseous 
product after the removal of the carbon dioxide (absorption by 
lime). Temperature and pressure a constant. Ans. 20 c.c. 

219. A mixture of 200 c.c. of methane and 300 c.c. of carbon 
monoxide was exploded with 600 c.c. of oxygen (excess). Cal¬ 
culate the volume of gaseous product. Calculate, also, the 
volume of product when deprived of aqueous vapor. Tem¬ 
perature of 100° and pressure of 760 mm. a constant throughout. 

Ans. 950 c.c. (moist). 

550 c.c. (dry). 

Note. — Since the oxygen is in excess we know the calculation ' 
is to be based upon the CH 4 and CO. The oxygen left unconsumed 
in these two reactions is readily determined by difference. 

220. A mixture of 80 c.c. of methane and 200 c.c. of oxygen 

was exploded over mercury. Calculate the volume of product. 
A temperature of 20° and barometric pressure of 757.4 mm. 
constant throughout. Ans. 122.83 c.c. 

Note. —The volume of dry gaseous product, at 20° and 757.4 mm., 
is simply to be changed to accord with the observed conditions in 
the presence of aqueous vapor. These experiments are usually carried 
out over mercury. The excess of water condenses of course upon 
the mercury column. If an appreciable quantity were present allow¬ 
ance for its pressure upon the column would need to be made. 

221. A mixture of 400 c.c. of hydrogen and 300 c.c. of 
oxygen was exploded. Calculate the volume of product. A 
temperature of 25° and pressure of 753.6 mm. constant. 

Ans. 103.23 c.c. 


COMBINATIONS BETWEEN GASES BY VOLUME 141 


222. A mixture of 200 c.c. of carbon monoxide and 300 c.c. 

of oxygen was exploded over mercury. Calculate the volume 
of product. A temperature of 27° and pressure of 740 mm. 
constant throughout. Ans. 400 c.c. 

223. A mixture of oxygen and hydrogen, measuring 150 c.c. 

in volume, was exploded by means of an electric spark. The 
volume of product measured 125 c.c. Temperature of 100° 
and atmospheric pressure a constant. Calculate the volumes 
of hydrogen and oxygen concerned in the reaction. The resid¬ 
ual gas is of course the excess of either the hydrogen or oxygen 
present. Ans. 50 c.c. hydrogen. 

25 c.c. oxygen. 

224. A mixture of oxygen and hydrogen, measuring 150 c.c. 

in volume, was exploded. The volume of product measured 
76.47 c.c. Temperature of 17° and pressure of 754.4 mm. a 
constant. Calculate the volumes of hydrogen and oxygen con¬ 
cerned in the reaction. Ans . 50 c.c. hydrogen. 

25 c.c. oxygen. 

225. 250 c.c. of dry air were mixed with 150 c.c. of hydrogen 

(an excess) and exploded. The volume of product was 350 c.c. 
Calculate the percentage of oxygen (by volume) fn the sample 
of air. Temperature of 100° and atmospheric pressure a con¬ 
stant. Ans. 20 per cent. 

226. 300 c.c. of dry air were mixed with 250 c.c. of hydrogen 

(an excess) and exploded. The volume of the dry gaseous 
product was 361 c.c. Calculate the percentage of oxygen (by 
volume) in the sample of air. Temperature of 20° and pressure 
of 745 mm. a constant. Ans. 21 per cent. 

221. A mixture of hydrogen sulphide with an excess of 
oxygen measured 350 c.c. at 100° and 750 mm. pressure. After 
explosion (with complete combustion) the volume of dry gase¬ 
ous product read 260 c.c. at these same conditions. Calculate 
the volume of hydrogen sulphide in the mixture. Ans. 60 c.c. 

228. A mixture of hydrogen sulphide with an excess of 
oxygen measured 200 c.c. at 20° and 747.4 mm. pressure. 
After explosion (with complete combustion) the volume of 
product read 143.3 c.c. at these same conditions. Calculate 
the volume of hydrogen sulphide in the mixture. Ans. 40 c.c. 


142 


CHEMICAL CALCULATIONS 


229. A mixture of acetylene, C 2 H 2 , with an excess of oxygen 

measured 350 c.c. at 25° and 745 mm. pressure. After explo¬ 
sion the volume of dry gaseous product read 275 c.c. at the 
same conditions. Calculate the volume of acetylene in the 
mixture. Ans. 50 c.c. 

230. A mixture of acetylene, C 2 H 2 , with an excess of oxygen 

measured 240 c.c. at 24° and 752.4 mm. pressure. After ex¬ 
plosion the volume of product read 221.8 c.c. at 28° and 750.1 
mm. pressure. Calculate the volume of acetylene in the mix¬ 
ture. Ans. 20 c.c. 

231. An excess of oxygen, 300 c.c., was admitted into a 
volume of ammonia (containing an impurity of air, i.e., nitro¬ 
gen and oxygen) which measured 240 c.c., and the mixture, at 
22° and 745.2 mm. pressure, exploded. The volume of product 
measured 415.1 c.c. at 24° and the same pressure. Calculate 
the percentage purity of the sample of ammonia. 

Ans. 80 per cent. 

232. Into a stoppered tube (90 c.c. in capacity) filled with 
chlorine, a small quantity of concentrated ammonia water 
(10 c.c. or an excess) was admitted. After shaking, the mouth 
of the tube was opened under a dilute acid solution contained 
in a tall cylinder. The excess of ammonia, together with the 
hydrogen chloride, formed in the reaction, were thus removed. 
The residual volume of gas may be considered as nitrogen. 
Calculate the volume of nitrogen in the tube. The room tem¬ 
perature and pressure considered a constant throughout. 

Ans. 30 c.c. 

233. A mixture of carbon monoxide and methane contain¬ 
ing also nitrogen measured 260 c.c. Into this mixture was 
admitted an excess of oxygen, 300 c.c., and the whole exploded. 
After the explosion the volume of product measured 520 c.c., 
but after the removal of aqueous vapor (by absorption) from the 
gaseous product the volume measured 320 c.c. Calculate the 
volume of each component in the original mixture. A tem¬ 
perature of 100° and atmospheric pressure constant. 

Ans. 80 c.c. CO. 

100 c.c. CH 4 . 

80 c.c. N. 


COMBINATIONS BETWEEN GASES BY VOLUME 143 


234. A sample of water gas containing an impurity of air 
measured 240 c.c. in volume. Into this was admitted an 
excess of oxygen, 400 c.c., and the mixture exploded. The 
volume of product measured 540 c.c. After removal of aque¬ 
ous vapor (by absorption) the volume of product measured 
440 c.c. Calculate the percentage of hydrogen and carbon 
monoxide in the sample. A temperature of 100° and atmos¬ 
pheric pressure constant throughout. 

Ans. 41.25 per cent H. 

41.25 per cent CO. 

235. 500 c.c. of a mixture of ammonia, methane and nitrogen 
were mixed with 500 c.c. of oxygen and the mixture exploded. 
The volume of the product measured 1050 c.c. A constant 
temperature of 100° and pressure of 760 mm. maintained 
throughout. Upon cooling the gaseous product to remove the 
aqueous vapor and recalculation of the volume of dry gas to 
the original conditions (100°), the volume measured 550 c.c. 
What was the composition of the original mixture? 

Ans. 100 c.c. CH 4 . 

200 c.c. NH 3 . 

200 c.c. N. 

236. A mixture of methane; hydrogen sulphide and air 

measured 300 c.c. When mixed with an excess of oxygen, 
400 c.c., and exploded the volume of product measured 620 
c.c. A temperature of 100° and atmospheric pressure constant 
throughout. Upon cooling the gaseous product to the room 
temperature and recalculation of the volume of dry gas to the 
original conditions, the volume measured 220 c.c. Calculate the 
composition of the original mixture. Ans. 120 c.c. CH 4 . 

160 c.c. H 2 S. 

20 c.c. air. 

237. A mixture of methane and acetylene measuring 500 
c.c. was mixed with 1500 c.c. of oxygen (an excess) and ex¬ 
ploded. The volume of product measured 1700 c.c. A tem¬ 
perature of 100° and atmospheric pressure constant. Upon 
removal of aqueous vapor (by condensation and recalcula¬ 
tion to the original conditions) the volume measured 1000 c.c. 
Calculate the composition of the original mixture. 

Ans. 200 c.c. CH 4 . 

300 c.c. C 2 H 2 . 


144 


CHEMICAL CALCULATIONS 


238. A mixture of hydrogen, methane and nitrogen measur¬ 
ing 350 c.c. was exploded with an excess of oxygen, 500 c.c. 
After the explosion the volume of dry gaseous product meas¬ 
ured 475 c.c. Upon removal of carbon dioxide the volume 
of gas measured 400 c.c. Calculate the composition of the 
original mixture. Room temperature and pressure a constant. 

Ans. 150 c.c. H. 

75 c.c. CH 4 . 
125 c.c. N. 

239. A mixture of cyanogen, methane and nitrogen meas¬ 
uring 200 c.c. in volume was exploded with an excess of oxygen, 
500 c.c. After the explosion the volume of product meas¬ 
ured 619.6 c.c. Upon removal of carbon dioxide (by lime) 
the volume of dry gas measured 430 c.c. Calculate the com¬ 
position of the original mixture. A temperature of 25° and 
pressure of 746.6 mm. constant throughout. 

Ans. 50 c.c. CH 4 . 

60 c.c. C 2 N a . 

90 c.c. N. 

240. A liter of methane contaminated with the vapor of 
carbon disulphide was mixed with an excess of oxygen, 3000 c.c., 
and exploded. The volume of product measured 3840 c.c. 
Calculate the percentage impurity of the methane. A tempera¬ 
ture of 100° and atmospheric pressure constant throughout. 

Ans. 16 per cent. 

241. An excess of oxygen was admitted to a vessel contain¬ 
ing hydrogen and ammonia and the mixture exploded. After 
the explosion there was observed no change in volume of 
product from that of the mixture. Calculate the percentage 
composition of the hydrogen and ammonia mixture. A tem¬ 
perature of 100° and atmospheric pressure constant. 

Ans. 33£ per cent H. 

66f per cent NH a . 

242. A mixture of ethylene and ammonia, contaminated 
with air measured 600 c.c. When this volume of gas was mixed 
with an excess of oxygen, 2000 c.c., and exploded, the volume 
of product measured 2640 c.c. Upon removal of aqueous 
vapor (by cooling and recalculation) the volume of dry gas 








COMBINATIONS BETWEEN GASES BY VOLUME 145 


measured 2120 c.c. Calculate the composition of the original 
mixture. A temperature of 100° and atmospheric pressure 
constant. Ans. 160 c.c. NH 3 . 

140 c.c. C 2 H 4 . 

300 c.c. air. 

243. A mixture of ethylene and ammonia contaminated 

with air measured 400 c.c. When this volume of gas was 
mixed with an excess of oxygen, 1200 c.c., and exploded, the 
volume of product measured 1293.6 c.c. Upon removal of 
carbon dioxide (by lime) the volume of dry gas measured 
1050 c.c. Calculate the composition of the original mixture. 
A temperature of 26° and pressure of 745.6 mm. constant 
throughout. Ans . 100 c.c. C 2 H 4 . 

120 c.c. NH 3 . 

180 c.c. air. 

244. Into a mixture of cyanogen; ethylene and nitrogen was 

admitted a large excess of oxygen and the mixture, which 
measured 1600 c.c. at 22° and 753 mm. pressure, exploded. 
After the explosion the volume of product measured 1257 c.c. 
at 24° and 746 mm. pressure. Upon removal of carbon dioxide 
and moisture the volume of dry gas (still at 24° and 746 mm.) 
measured 609.7 c.c. Calculate the amount of cyanogen and 
ethylene in the original mixture. Ans. 100 c.c. C 2 N 2 . 

200 c.c. C 2 H 4 . 

245. A mixture of hydrogen and ammonia contaminated 

with air measured 400 c.c. Into this was admitted an excess 
of oxygen, 600 c.c., and the final mixture exploded. After the 
explosion the volume of product measured 1025 c.c. Upon 
removal of aqueous vapor (by absorption) the volume read 
675 c.c. Calculate the composition of the original mixture. A 
temperature of 100° and atmospheric pressure constant through¬ 
out. (Cf. Ex. 63.) Ans. 50 c.c. H. 

200 c.c. NH 3 . 

150 c.c. air. 

246. A mixture of hydrogen, methane, carbon monoxide 
and nitrogen measured 500 c.c. Into this was admitted an 
excess of oxygen, 900 c.c., and the final mixture exploded. 
After the explosion the volume of product measured 1250 c.c. 
Upon removal of aqueous vapor (by absorption) the volume 
measured 1000 c.c. Upon removal of carbon dioxide (by 


146 


CHEMICAL CALCULATIONS 


lime) the volume measured 725 c.c. Calculate the composition 
of the original mixture. A temperature of 100° and atmos¬ 
pheric pressure constant. (Cf. Ex. 63.) Am. 100 c.c. H. 

200 c.c. CO. 

75 c.c. CH 4 . 

125 c.c. N. 

347. A mixture of carbon monoxide and methane contami¬ 
nated with air and measuring 400 c.c. was exploded with an 
excess of oxygen, 1000 c.c. The volume of product measured 
1025.4 c.c. Upon removal of the carbon dioxide (by lime) the 
volume of dry gas measured 725 c.c. Calculate the composi¬ 
tion of the original mixture. A temperature of 21° and pres¬ 
sure of 745.5 mm. constant throughout. Am. 100 c.c. CO. 

(Cf. Ex. 63.) 175 c.c. CH 4 . 

125 c.c. air. 

348. A mixture of cyanogen, methane and hydrogen con¬ 

taminated with air measured 800 c.c. An excess of oxygen, 
2000 c.c., was admitted into the vessel containing this mixture 
and the final mixture exploded. After the explosion the vol¬ 
ume of product measured 2750 c.c. After removal of aqueous 
vapor (by cooling and recalculation) the volume measured 2250 
c.c. After removal of carbon dioxide (by lime) the final vol¬ 
ume measured 1450 c.c. Calculate the composition of the 
original mixture. A temperature of 100° and atmospheric 
pressure constant. Am. 100 c.c. H. 

(Cf. Ex. 63.) 200 c.c. CH 4 . 

300 c.c. C 2 N 2 . 

100 c.c. air. 

349. An excess of oxygen, 800 c.c., was added to a mixture 

of ammonia, ethylene and hydrogen, 310 c.c. in volume, and 
the final mixture exploded. After the explosion the volume 
of product measured 648.35 c.c. Upon removal of carbon 
dioxide (by lime) the volume of dry gas measured 435 c.c. 
Calculate the composition of the original mixture. A tempera¬ 
ture of 18° and pressure of 746 mm. constant throughout. 
(Cf. Ex. 63.) Am. 50 c.c. H. 

100 c.c. C 2 H 4 . 

160 c.c. NH a . 

Note. —The contraction due to the reactions is here masked 
in the contraction from the loss of aqueous vapor. Construct the 


COMBINATIONS BETWEEN GASES BY VOLUME 147 


third equation from the volume of oxygen required in the several 
combustions, bearing in mind that the volume, 365 c.c. (800 c.c. — 
435 c.c.), which nominally represents this quantity in the problem, 
is in fact less than the actual volume by that volume of nitrogen left 
free. 

250. A mixture of methane and ethylene measuring 300 c.c. 

was exploded with an excess of oxygen, 1500 c.c. After the 
explosion the volume of dry gaseous product (calculated to 
observed conditions of temperature and pressure) measured 
1200 c.c. In order to determine the volume of oxygen actually 
consumed in this combustion and thus present data upon which 
a second equation may be constructed, an excess of hydrogen, 
900 c.c., was admitted to the vessel containing the dry gaseous 
product. After an explosion the volume of this second product 
measured 1400 c.c. (calculated as above, with absence of 
aqueous vapor). Calculate the amount of methane and ethyl¬ 
ene in the original mixture. The room temperature and pres¬ 
sure remained constant throughout. Ans. 200 c.c. CH 4 . 
(Cf. Ex. 63.) 100 c.c. C 2 H 4 . 

251. A volume of nitric oxide, NO, measuring 400 c.c. was 

required for the combustion of a definite weight of phosphorus. 
What volume of nitrogen remained free? The room tempera¬ 
ture and pressure a constant. Ans. 200 c.c. 

252. A quantity of phosphorus was burned in a vessel 
over water holding 222.7 c.c. of nitrous oxide, N 2 0, meas¬ 
ured at 20° and 750 mm. pressure. What volume of nitrogen 
remained, the temperature and pressure constant? 

Ans. 222.7 c.c. 

253. What volume of water gas is theoretically possible 

from the action of 1 liter of steam upon heated coke? Tempera¬ 
ture and pressure a constant. Ans. 2000 c.c. 

254. In the decomposition of methane by chlorine what 
volume of hydrogen chloride corresponds to 1 volume of meth¬ 
ane? Temperature and pressure constant. Ans. 4 volumes. 

255. A mixture of 400 c.c. of methane and 1000 c.c. of chlo¬ 
rine was exploded. Calculate the volume of gaseous product, 

j temperature and pressure a constant. Ans. 1800 c.c. 

256. Wliat volume of gaseous product may be obtained in 
the decomposition of 100 c.c. of ammonia by heated cupric 
oxide? Temperature and pressure a constant. 

Ans. 200 c.c. 

(Steam + N). 



CHAPTER XII. 


COMPLEX EQUATIONS. 

The study of chemical reactions as outlined in Chapter 
IX was not inclusive of those examples where a change 
in the valence of an element occurs. Although these 
changes may complicate the matter of drawing up the 
equations, the same principles as already discussed will 
be found to apply. The definite quantity of each element 
concerned, whether or not it exerts the same combin¬ 
ing capacity for other elements on the two sides of an 
equation, is nevertheless a constant for the particular 
equation in question. The exact application of these 
principles will rest, of course, upon the construction of the 
equations in their molecular form. 

Oxidation and Reduction. — The increase in valence or 
combining capacity* which accompanies an element when, 
for instance, it passes from a lower to a higher oxide, has 
been taken as the measure of the degree of oxidation. 
The position of oxygen in this connection might be 
filled by almost any non-metallic element and still we 
may have the means of measuring the increase in com¬ 
bining capacity of a metallic element or radical acting as 
such toward a non-metallic element, i.e., oxidation. The 
unit of measure, rather than oxygen with its 2 com¬ 
bining capacities, is referred to an element with but 1 

* The author is indebted to his colleague, Professor S. Laurence 
Bigelow, for this interpretation of valence as the capacity factor 
of chemical energy, i.e., the combining capacity of a unit quantity 
of an element or radical. 


148 


COMPLEX EQUATIONS 


149 


capacity, e.g. hydrogen. Thus 1 molecule of chromic 
oxide, Cr 2 nl 0 3 n , undergoes oxidation to 2 molecules of 
chromium trioxide, 2 Cr VI 0 3 n , and the oxygen intake for 
this definite quantity of chromium is 3 oxygen atoms, 
i.e., the equivalent of 6 hydrogen atoms in combining 
capacity, or 6 combining capacities. This amount for 
2 atoms of chromium is equivalent to 3 combining capac¬ 
ities for 1 atom of chromium and thus this increase will 
be denoted in the second compound Cr VI 0 3 n . 

The reverse process, or a decrease in this combining 
capacity toward a non-metallic element, is known as 
reduction. For example, ferric iron as Fe 111 ^ 1 is reduced 
to ferrous iron, Fe lx Ci 2 l . 

Analogous to this the increase in combining capacity of 
a non-metallic element, or a radical acting as such, for 
oxygen or some other non-metal will be a measure of the 
oxidation, e.g. C^O 11 —> C1 2 VII 0 7 11 . When, however, this 
capacity of a non-metal is considered in relation to its 
combination with metallic groups the increase in its com¬ 
bining capacity is in fact a measure of its reduction. 
Thus potassium ferricyanide, K 3 i rFe ra (CN) (} 1 ] 111 , is reduced 
to potassium ferrocyanide, K 4 I [Fe 11 (CN) 6 I ] IV , when the ferri¬ 
cyanide radical increases its combining capacity toward 
the metallic element potassium. This result is of course 
due to the reduction of the ferric iron in the former 
radical to ferrous iron in the latter. In general the 
increase in the combining capacity of a non-metallic 
radical for a metallic one signifies a demand for a lower 
number of these capacities required of an intra-radical 
metal for complete combination, or what was previously 
interpreted as a reduction. 

Reaction-Quantities Involved in Oxidation and Reduc¬ 
tion. — The change of an element from the non-metallic 
to the metallic groupings, or vice versa, may be accom- 


150 


CHEMICAL CALCULATIONS 


panied by a loss or gain in its combining capacity, and 
consequently we speak of it as undergoing a reduction or 
oxidation. Thus potassium dichromate, K 2 Cr 2 0 7 , con¬ 
tains the chromium associated directly with oxygen in 
the form of the acid H 2 Cr0 4 (or oxide Cr VI 0 3 n ), but by the 
action of a strong acid this chromium, with a combining 
capacity of 6, is completely transformed into a chromic 
compound (where chromium is the basic element) and then 
has a combining capacity of only 3, Cr 2 III 0 3 11 . The action 
therefore is described as one of reduction. 

In illustration of these points a few of the more com¬ 
mon examples will be given. First we may mention the 
action of hydrogen sulphide upon a solution of ferric 
chloride: 

2 FeCl 3 + H 2 S = 2 FeCl 2 + 2 HC1 + S. 
Expressed by ions: 

2 Fe‘" + 6 Cl' + 2 H* + S" 

= 2 Fe" + 4 Cl' +2 H’ + 2 Cl' + S/ J 

where the combining capacities are represented in solu- 
tion as ionic charges. 

From the reaction-quantities noted we may draw up 
the several ratios of proportionality here existent. For j 
example the amount of ferrous chloride resulting from a 
definite weight of ferric chloride is expressed by the 
ratio FeCl 2 /FeCl 3 , and then again the weight (as well as 
the volume eventually) of hydrogen sulphide necessary 
for this reduction of the ferric chloride is in accord with 
the ratio H 2 S/2 FeCl 3 . The hydrogen sulphide molecule 
is equivalent therefore in its reducing action to 2 units 
of hydrogen, whereas 2 molecules of ferric chloride are 
equivalent in oxidizing action to 1 oxygen atom. 

Oxidizing Action of Potassium Dichromate. — As has 
been stated, the oxidizing action of potassium dichro- 


COMPLEX EQUATIONS 


151 


mate, K 2 Cr 2 0 7 , in the presence of a strong acid and sub¬ 
stance capable of oxidation, is due to the transformation 
of the dichromate (Cr VI ) into a chromic compound (Cr in ). 
The oxidizing power of this substance is measured by 
the liberation of 3 atoms of oxygen per molecule of 
dichromate, as may be shown by the following hypotheti¬ 
cal equation for the action with sulphuric acid: 

K 2 Cr 2 0 7 + 4 H 2 S0 4 = + Cr 2 (S0 4 ) 3 + 4 H 2 0 + 3 O. 

When an active reducing agent, e.g. ammonium sul¬ 
phide, is present the reduction of the dichromate may be 
accomplished without the presence of an acid: 

K 2 Cr 2 0 7 + 3 (NH 4 ) 2 S + 7 H 2 0 

= 2 Cr(OH) 3 + 2 KOH + 6 NH 4 OH + 3 S. 

In other cases some oxidizable substance must be added 
to the acid mixture in order that the oxygen capable of 
liberation may be actually consumed and the reaction 
proceed. 

A number of substances, such as hydrogen sulphide, 
sulphur dioxide, carbon monoxide and alcohol, are readily 
oxidized in this manner to sulphur, sulphur trioxide, 
carbon dioxide and aldehyde, respectively. Thus with 
hydrogen sulphide: 

K 2 Cr 2 0 7 + 4 + 3 

= + Cr 2 (S0 4 ) 3 -f 4 H 2 0 + 3 H 2 0 + 3 S. 

The presence of 3 molecules of hydrogen sulphide is neces¬ 
sary for interaction with the 3 atoms of oxygen liberated 
per molecule of dichromate; the products by this oxidation 
are indicated in the equation. 1 molecular weight of 
potassium dichromate is here directly proportional to 
3 molecular weights of hydrogen sulphide, hence in this 


152 


CHEMICAL CALCULATIONS 


reaction all values for these substances must accord with 
the ratio 

3 3 (34.09) 

KjOjO, ’ ° r 294.2 ' 

From the relation between the molecular weight of a 
substance and its volume as a vapor we are able to calcu¬ 
late, for example, the volume of hydrogen sulphide that 
can be oxidized by a corresponding proportional weight of 
the dichromate. 

Example 66. — What volume of hydrogen sulphide, at stand¬ 
ard conditions, can be oxidized by 7.36 grams of potassium 
dichromate in an acid solution? 

The volume of hydrogen sulphide oxidized by 1 gram- 
molecular weight of potassium dichromate (294.2 grams) 
is that volume corresponding to 3 gram-molecular weights 
of the gas, or 67,200 c.c. ( i.e ., 3 X 22,400 c.c.) at standard 
conditions. The direct proportionality between the gram- 
molecular weight and the actual weight of dichromate 
here given, determines also the proportionality between 
the respective volumes of hydrogen sulphide correspond¬ 
ing to these values: 

294.2 : 7.36 = 67,200 : x, 

in which x represents the volume (1681 c.c.) of gas oxi¬ 
dized. 

The weight of sulphur precipitated, as well as any other 
substance present, may be calculated from the standpoint 
of these proportional values expressed in the equation. 
For example, the relation between the sulphur and potas¬ 
sium dichromate is expressed by the ratio: 

3 S 96.21 

K 2 Cr 2 <V 294.2 ‘ 






COMPLEX EQUATIONS 


153 


When hydrochloric acid is used the excess of acid is 
itself oxidized by the liberated oxygen with the produc¬ 
tion of free chlorine and water: 

K 2 Cr 2 0 7 + 14 HC1 = 2 KC1 + 2 CrCl 3 + 7 H 2 0 + 3 Cl 2 . 

Standard Oxidizing Solutions with Potassium Dichro¬ 
mate. — From reactions of oxidation with a potassium 
dichromate solution we are able to draw up the value of 
such a solution in terms of the hydrogen equivalent. 
1 atom of oxygen will oxidize and combine with 2 atoms 
of hydrogen, hence the equivalence of 3 atoms of oxygen 
in terms of hydrogen is necessarily 6. Since 1.01 grams 
of hydrogen or this unit equivalent are required per liter 
of a normal solution, we shall require here in a normal solu¬ 
tion of potassium dichromate—that solution which per liter 
can furnish just sufficient oxygen to unite with 1.01 grams 
of hydrogen — exactly one-sixth of the gram-molecular 
weight of this salt, i.e., 294.2/6 or 49.03 grams. In the 
work of titration with standard solutions of this salt the 
change in color from red to green indicates complete 
reduction, but for accurate determination of this point 
it is customary to involve small portions of the solution 
in further reactions. 

Example 67. — What is the normality of that potassium 
dichromate solution, 40 c.c. of which oxidized 0.590 gram of 
ferrous sulphate to the ferric state? 

The more commonly used ferrous ammonium sulphate, 
FeS0 4 (NH 4 )2S0 4 .6 H 2 0, is here replaced by ferrous 
sulphate, FeS0 4 , for the sake of simplicity. The reaction 
for the oxidation of this latter to ferric sulphate involves 
1 oxygen atom, with 1 molecule of sulphuric acid, per 2 
molecules of the ferrous salt: 

2 FeS0 4 + H 2 S0 4 + O = Fe 2 (S0 4 ) 3 + H 2 0. 


154 


CHEMICAL CALCULATIONS 


Consequently to include this action in the equation which 
represents the oxidation by means of 1 molecule of potas¬ 
sium dichromate, with its 3 possible oxygen atoms, we 
shall need to multiply the equation above by 3 and thus 
obtain 3 0 as a reaction-quantity common to both equa¬ 
tions : 

K 2 Cr 2 0 7 + 4 H 2 S0 4 = K 2 S0 4 + Cr 2 (S0 4 ) 3 + 4 H 2 0 + 30 
6 FeS0 4 + 3 H 2 S0 4 + 30 = 3 Fe 2 (S0 4 ) 3 + 3 H 2 0. 

It comes to the same end if we simply combine these 
equations and let this common quantity (3 O) drop out, 
when we shall have the final expression: 

K 2 Cr 2 0 7 + 6 FeS0 4 + 7 H 2 S0 4 = 

K 2 S0 4 + Cr 2 (S0 4 ) 3 + 3 Fe 2 (S0 4 ) 3 + 7 H 2 0. 

In either case the reaction-quantities 6 FeS0 4 and K 2 Cr 2 0 7 
are directly proportional to each other, and the ratio 
6 FeS0 4 /K 2 Cr 2 0 7 , or 911.52/294.2, determines the ratio 
between all weights which severally may represent these 
quantities. 

In the example 0.590 gram of FeS0 4 was oxidized. 
Therefore the following equivalence in the ratios: 

911.52 = 0.590 
294.2 x 

The value for x is found to be 0.1904 gram. This weight, 
however, was present in 40 c.c. of solvent, consequently 
in 1 liter we should have 25 X 0.1904 or 4.76 grams of 
potassium dichromate. A normal solution of this salt 
contains 49.03 grams per liter (p. 153), hence we have 
a solution of 4.76/49.03 normality, i.e., 0.0971 normal, a 
trifle less than N/10. 

Oxidizing Action of Potassium Permanganate. — As a 

second illustration the oxidizing action of potassium per- 



COMPLEX EQUATIONS 


155 


manganate is of much importance. The action in an 
acid solution is analogous to that of potassium dichromate, 
only the reduction of Mn vn to Mn 11 is accompanied here 
with the liberation of 5 atoms of oxygen for every 2 mole¬ 
cules of permanganate: 

2 KMn0 4 + 3 H 2 S0 4 = K 2 S0 4 + 2 MnS0 4 + 3 H 2 0 + 5 0. 

The presence of some substance capable of oxidation is 
necessary in order to bring about this reaction in dilute 
solutions. We may take again the example of hydrogen 
sulphide as affected by an oxidizing agent: 

2 KMnO, + 3 H 2 S0 4 + 5 H 2 S 

= K 2 S0 4 + 2 MnS0 4 + 3 H 2 0 + 4 H 2 0 + 5 S. 

From this equation the relative amounts of hydrogen sul¬ 
phide and potassium permanganate involved will always 
accord with the ratio between their respective reaction- 
quantities: 

5 H 2 S 5 (34.09) 

2 KMn0 4 ’ 2 (158.03)* 

When expressed by molecular equations the substances 
here concerned may be studied also in regard to their 
volume relations. 

Example 68. — What weight of potassium permanganate will 
be reduced by 2800 c.c. of hydrogen sulphide, at standard condi¬ 
tions of temperature and pressure? 

The reaction already discussed supplies the necessary 
data: 2 gram-molecular weights (316.06 grams) of the 
permanganate are reduced by 5 X 22,400 c.c. of hydrogen 
sulphide. By simple proportion the weight of perman¬ 
ganate that can be reduced by 2800 c.c. is derived as 
follows: 

(5 X 22.400) : 2800 = 316.06 : x, 
where x, or 7.9 grams, is this weight of permanganate. 




156 


CHEMICAL CALCULATIONS 


The reverse operation, or that of determining the weight 
(or volume) of hydrogen sulphide oxidized by a known 
weight of potassium permanganate, follows the same 
principle. 

Standard Oxidizing Solutions with Potassium Perman¬ 
ganate. — The liberation of 5 atoms of oxygen per 2 
molecules of permanganate also determines the amount 
of this salt that must be present in one liter of its normal 
solution for use in the presence of acid. By calculation, 
in a manner similar to that described under the standard 
dichromate solution, it is found necessary to have one- 
tenth of the 2 gram-molecular weights of potassium 
permanganate (316.06 grams), or 31.606 grams, in each 
liter of solution; whereby this volume of solution will be 
capable of furnishing just sufficient oxygen to oxidize 
1.01 grams of hydrogen. The change from the deep 
purple color of the permanganate to the almost colorless 
manganous salt serves admirably to mark the point when 
the reduction is complete and hence to fix the end-point 
in titration with a permanganate solution. 

Example 69. — 100 c.c. of a hydrogen peroxide solution were 
required to decolorize 500 c.c. of N/10 potassium permanganate 
solution (acidulated). What was the percentage concentration 
of the hydrogen peroxide solution? 

2 KMn0 4 + 3 H 2 S0 4 + 5 H 2 0 2 

= K 2 S0 4 + 2 MnS0 4 + 3 H 2 0 + 5 H 2 0 + 5 0 2 . 

As the equation indicates, the 5 atoms of oxygen liberated 
from the permanganate withdraw 5 atoms of oxygen from 
5 molecules of hydrogen peroxide and are together removed 
from the reaction in the molecular form (5 molecules). 
500 c.c. of N/10 potassium permanganate solution are 
equivalent to 50 c.c. of the normal solution. One liter of 


COMPLEX EQUATIONS 


157 


a normal potassium permanganate solution, containing 
31,606 grams (1/10 of 316.06 grams as represented by the 
reaction-quantity), can only react with 1/10 of the quan¬ 
tity 5 H 2 0 2 required by the reaction, or with 1/2 of one 
molecule of hydrogen peroxide. Here 50 c.c. (1/20 liter) 
can act upon just 1/20 of the amount of hydrogen per¬ 
oxide that a liter of the normal solution can oxidize 
(namely, 1/2 of a gram-molecular weight, or 17 grams), 
i.e. } 1/20 of 17 grams or 0.85 gram. Consequently in 
100 c.c. of the hydrogen peroxide solution there is only 
this amount, 0.85 gram, of the compound, and if the 
weight of the solution is assumed to be 100 grams, then 
the percentage of hydrogen peroxide present is 85/100 of 

1 per cent. 

From the conditions of this reaction the amount of any 
one substance present can be calculated also from the 
volume of oxygen evolved. 

Just as potassium dichromate so also an acidulated 
solution of potassium permanganate readily oxidizes solu¬ 
tions of ferrous salts to the ferric state. The equation 
indicates, of course, this liberation of 5 oxygen atoms per 

2 molecules of permanganate: 

2 KMn0 4 + 10 FeS0 4 + 8 H 2 S0 4 

= K 2 S0 4 + 2 MnS0 4 + 5 Fe 2 (S0 4 ) 3 + 8 H 2 0. 

Again the action of an acidulated potassium perman¬ 
ganate solution upon a hot solution of oxalic acid (weighed 
out as C 2 H 2 0 4 .2 H 2 0), with the complete oxidation of 
the latter into carbon dioxide and water, affords an 
admirable means for the titration and standardization of 
permanganate solutions: 

2 KMn0 4 + 5 C 2 H 2 0 4 .2 H 2 0 + 3 H 2 S0 4 

= K 2 S0 4 + 2 MnS0 4 + 10 C0 3 + 8 H 2 0 + (10 H,0). 



158 


CHEMICAL CALCULATIONS 


Hydrochloric acid, except in the cold and under proper 
precautions, is rarely ever used for the acidulation of 
permanganate solutions owing to its ready oxidation, 
similar to that by potassium dichromate, into chlorine 
and water: 

2 KMn0 4 + 16 HC1 = 2 KC1 + 2 MnCl 2 + 5 Cl 2 + 8 H 2 0. 

In neutral or alkaline solutions potassium permanganate 
is reduced by the presence of oxidizable substances to 
manganese dioxide (precipitated), and somewhat less 
oxygen per molecule of permanganate is thereby liberated. 
The action may be represented in part as follows: 

2 KMn0 4 + H 2 0 = 2 KOH + 2 Mn0 2 + 3 0 . 

The 3 oxygen atoms liberated from 2 molecules of potas¬ 
sium permanganate determine the amount of this salt 
in its normal solution to be 1/6 of 316.06 grams (repre¬ 
sented by the reaction-quantity 2 KMn0 4 ), or 52.67 
grams per liter. 

Iodimetry. — The action of iodine in the presence of 
water upon a number of compounds is interesting to show 
one of the indirect methods of estimating substances in 
solution. 

Iodine dissolves readily in water containing potassium 
iodide. This solution acts as a mild oxidizing agent as 
may be seen from the following equation: 

S0 2 + 2 H 2 0 + I 2 = H 2 S0 4 + 2 HI. 

In the presence of an oxidizable substance the one mole¬ 
cule of iodine, through the decomposition of water present, 
furnishes sufficient oxygen to oxidize one molecule of 
sulphur dioxide to the trioxide, or an amount capable 
also of oxidizing two atoms of hydrogen. The normal 
value, therefore, of such an iodine solution must be fixed 


COMPLEX EQUATIONS 


159 


at one-half the gram-molecular weight (253.84 grams), or 
126.92 grams per liter. The employment, however, of a 
more dilute solution as N/10, etc., is more common in 
practice. The end-point in titrations with iodine solu¬ 
tions is detected by the addition of a slight amount of a 
starch solution, with which substance the iodine, when 
occurring in slight excess, forms a deep blue mixture. 

An iodine solution may also exert an oxidizing action 
by withdrawing a number of positively charged ions 
from a salt in solution: 

2 Na 2 S 2 0 3 + I 2 = Na 2 S 4 0 6 + 2 Nal, or 

4 Na + 2 S 2 0 3 " + I 2 = 2 Na* + S 4 0" + 2 Na* + 2 I'. 

A solution of sodium thiosulphate can be easily made up 
from the well crystallized salt, Na 2 S 2 0 3 .5 H 2 0, and this 
solution titrated against an iodine solution, using starch 
as an indicator. One molecule of iodine reacts with two 
molecules of the thiosulphate; consequently a normal 
solution of the latter will contain (with reference to a 
normal solution of iodine) just one-half of this reaction- 
quantity or one gram-molecular weight per liter. 

Substances that are capable of liberating iodine from 
potassium iodide, e.g. chlorine, bromine, etc., may be 
brought in this way into estimations by volumetric 
analysis: 

2 KI + Cl 2 = 2 KC1 + I 2 . 

The amount of iodine liberated is in exact accord with 
the proportional amount of chlorine under investigation, 
and the iodine in turn is open to determination from the 
amount of thiosulphate required for titration. This in¬ 
teresting application of volumetric analysis is known as 
Iodimetry . 

Example 70. — An unknown weight of sodium dichromate 
was warmed with an excess of hydrochloric acid, and the chlorine, 


160 


CHEMICAL CALCULATIONS 


set free, passed into a solution of potassium iodide. 400 c.c. of 
N/10 sodium thiosulphate solution were required for the titra¬ 
tion of the iodine thereby liberated. Determine the amount of 
chlorine evolved, and also the weight of dichromate reduced. 

400 c.c. of N/10 sodium thiosulphate solution are 
equivalent to 40 c.c. of a normal solution. 1 liter of a 
normal solution of this salt exactly decolorizes 1 liter of 
a normal solution of iodine, or this amount of iodine 
(126.92 grams) in whatever volume of solution it is con¬ 
tained. Consequently 40 c.c. of the normal thiosulphate 
solution will act upon that amount of iodine as indi¬ 
cated in the proportion: 

1000 : 40 = 126.92 : x, or x = 5.177 grams. 

The amount of chlorine proportional to this weight of 
iodine is in accord with the ratio Cl/1, or 35.46/126.92, 
as indicated by the reaction above. By proportion we 
have: 126.92 : 35.46 = 5.177 : x, where x } or 1.45 
grams, is the weight of chlorine set free apd thus 
made capable of liberating the amount of iodine above 
calculated. The determination of the amount of sodium 
dichromate necessary to liberate this weight of chlorine 
rests upon an equation exactly analogous to the action 
of hydrochloric acid upon potassium dichromate (see 
p. 153), and is indicated by the ratio: 

3 Cl, or 212.76 
Na 2 Cr 2 0 7 ’ 262 

Accordingly, 212.76 : 262 = 1.45 : x. From this we cal¬ 
culate the weight of dichromate as 1.786 grams. 

No matter how complex the equations may become, the 
true relationship between all of the substances concerned 
is found to be in direct proportionality to their respec¬ 
tive reaction-quantities. 




COMPLEX EQUATIONS 


161 


PROBLEMS. 

257 . What weight of ferric chloride, FeCl 3 , can be reduced 
to ferrous chloride, FeCl 2 , by 8.52 grams of hydrogen sulphide? 

Arts. 81.10 grams. 

258 . What volume of hydrogen sulphide (at standard con¬ 

ditions) will be required for the reduction of 100 grams of ferric 
chloride, FeCl 3 , to the ferrous salt? Ans. 6903.8 c.c. 

259 . Calculate the weight of sulphur precipitated in the 
reduction of 100 grams of ferric chloride, FeCl 3 , to the ferrous 
salt by the action of hydrogen sulphide. Ans. 9.884 grams. 

260. 1000 c.c. of hydrogen sulphide (at standard conditions) 

will reduce what weight of potassium dichromate in acid solu¬ 
tion? Ans. 4.378 grams. 

261 . What volume of hydrogen sulphide (at standard con¬ 

ditions) will be required for the reduction of 100 grams of potas¬ 
sium dichromate in acid solution? Ans. 22,842 c.c. 

262. What weight of hydrogen sulphide can be oxidized by 
400 grams of potassium dichromate, in acid solution? What 
weight of sulphur will be precipitated? 

Ans. 139.05 grams HjS. 

130.81 grams S. 

263 . What volume of sulphur dioxide (at standard condi¬ 

tions) can be oxidized by 30 grams of potassium dichromate in 
acid solution? Ans. 6852 c.c. 

264 . What volume of hydrogen sulphide, at 24° and 750 mm. 

pressure, can be oxidized by 10 grams of potassium dichromate 
in acid solution? Ans. 2518 c.c. 

265 . What weight of potassium dichromate, in acid solution, 

will be reduced by 1653.5 c.c. of sulphur dioxide at 22° and 
745 mm. pressure? Ans. 6.567 grams. 

266 . Calculate the volume of chlorine liberated in the action 

of 40 grams of potassium dichromate upon a hydrochloric acid 
solution. Ans. 9137 c.c. 

267 . What weight of sodium dichromate must enter into 

reaction with a hydrochloric acid solution in order to liberate 
100 grams of chlorine? Ans. 123.14 grams. 


162 


CHEMICAL CALCULATIONS 


268. Determine the normality of a potassium dichromate 

solution, 25 c.c. of which oxidized 1.24 grams of ferrous sulphate 
to the ferric salt. 0.3265 N. 

269. Determine the normality of a sodium dichromate solu¬ 
tion, 50 c.c. of which oxidized 3.85 grams of ferrous ammonium 
sulphate, FeS0 4 . (NH 4 ) 2 S0 4 .6 H 2 0 to the feme salt. 

Ans. 0.1963 N. 

270. What volume of hydrogen sulphide, at standard con¬ 

ditions, will be required to reduce 200 c.c. of N/10 potassium 
dichromate solution (acidulated)? Ans. 224 c.c. 

271. What weight of potassium permanganate, in acid solu¬ 

tion, will be reduced by 5000 c.c. of hydrogen sulphide at 
standard conditions? Ans. 14.11 grams. 

272. Calculate the volume of hydrogen sulphide, at stand¬ 

ard conditions, that can be oxidized by 4 grams of potassium 
permanganate in acid solution. Ans. 1417.4 c.c. 

273. Wliat weight of sulphur dioxide can be oxidized by 
200 grams of potassium permanganate in acid solution? 

Ans. 202.7 grams. 

274. 50 c.c. of a solution of hydrogen peroxide were required 

to decolorize 400 c.c. of N/5 potassium permanganate solution 
(acidulated). Calculate the percentage concentration of the 
hydrogen peroxide solution. Ans. 2.72 per cent. 

275. Calculate the weight of crystallized oxalic acid, C 2 H 2 0 4 . 

2 H 2 0, required for the reduction of 100 grams of potassium 
permanganate in acid solution. What volume of carbon diox¬ 
ide, at standard conditions, would be liberated? 

Ans. 199.4 grams. 

70,873 c.c. C0 2 . 

276. 50 c.c. of an acidulated potassium permanganate solu¬ 
tion were reduced by 2.4 grams of anhydrous oxalic add, 
C 2 H 2 0 4 . Calculate the normality of the permanganate solution, i 

Ans. 1.066 N. 

277. Wliat weight of sulphur dioxide will be oxidized by 50 
grams of potassium permanganate in alkaline solution? 

Ans. 3.04 grams. 




COMPLEX EQUATIONS 


163 


278. The bromine set free by the action of manganese 
dioxide upon a hydrobromic acid solution was passed into a 
solution of potassium iodide. 200 c.c. of N/10 sodium thio¬ 
sulphate solution were required for the titration of the free 
iodine. Calculate the weight of bromine evolved. 

Arts. 1.6 grams. 

279. Determine the purity of a sample of manganese dioxide, 
2.2 grams of which, with excess of hydrochloric acid, set free 
sufficient chlorine to liberate a quantity of iodine that required 
250 c.c. of N/5 sodium thiosulphate solution for titration. 

Ans. 98.79 per cent pure. 

280. 0.2452 gram of potassium dichromate, acting upon 

an excess of hydrochloric acid, set free an amount of chlorine 
which, when passed into a solution of potassium iodide, liber¬ 
ated iodine sufficient for the titration of 50 c.c. of an unknown 
sodium thiosulphate solution. Calculate the normality of this 
latter solution. Ans, N/10. 




APPENDIX. 










APPENDIX I. 

Correction of Barometic Readings. 

A barometric reading at room temperature is reduced 
to the corresponding height of a column of mercury at 0° 
by subtracting from the actual reading in millimeters 
that number in the second column below (Correction) set 
opposite the observed temperature. The barometric read¬ 
ings throughout these problems are given in corrected 
form. 


Tempera¬ 

ture. 

Correc¬ 

tion. 

Tempera¬ 

ture. 

Correc¬ 

tion. 

Tempera¬ 

ture. 

Correc¬ 

tion. 

12 

1.6 

17 

2.2 

23 

3.0 

13 

1.7 

18.5 

2.4 

24.5 

3.2 

14 

1.8 

20 

2.6 

25 

3.3 

15 

2!0 

21.5 

2.8 

26 

3.4 


APPENDIX II. 

Tension of Aqueous Vapor in Millimeters. 


Tempera¬ 

ture. 

Pressure. 

Tempera¬ 

ture. 

0° 

4.6 

16° 

5 

6.5 

17 

8 

8.0 

18 

9 

8.6 

19 

10 

9.2 

20 

11 

9.8 

21 

12 

10.5 

22 

13 

11.2 

23 

14 

11.9 

24 

15 

12.7 

25 


Pressure. 

Tempera¬ 

ture. 

Pressure. 

13.5 

26° 

25.1 

14.4 

27 

26.5 

15.4 

28 

28.1 

16.3 

29 

29.8 

17.4 

30 

31.5 

18.5 

31 

33.4 

19.7 

32 

35.4 

20.9 

33 

37.4 

22.2 

34 

39.6 

23.6 

100 

760.0 


167 





























168 


CHEMICAL CALCULATIONS 


APPENDIX III. — Four-Figure Logarithms. 


o 

^ CO 
<u 

Third figure. 

Fourth-figure 

differences. 

Em 

£« 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

4 

8 

12 

17 

21 

25 

29 

33 

37 

11 

0414 

0453 

0492 

0531 

0569 

0607 

0645 

0682 

0719 

0755 

4 

8 

11 

15 

19 

23 

26 

30 

34 

12 

0792 

0828 

0864 

0899 

0934 

0969 

1004 

1038 

1072 

1106 

3 

7 

10 

14 

17 

21 

24 

28 

31 

13 

1139 

1173 

1206 

1239 

1271 

1303 

1335 

1367 

1399 

1430 

3 

6 

10 

13 

16 

19 

23 

26 

29 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

1703 

1732 

3 

6 

9 

12 

15 

18 

21 

24 

27 

15 

1761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

3 

6 

8 

11 

14 

17 

20 

22 

25 

16 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

3 

5 

8 

11 

13 

16 

18 

21 

24 

17 

2304 

2330 

2355 

2380 

2405 

2430 

2455 

2480 

2504 

2529 

2 

5 

7 

10 

12 

15 

17 

20 

22 

18 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

2 

5 

7 

9 

12 

14 

16 

19 

21 

19 

2788 

2810 

2833 

2856 

2878 

2900 

2923 

2945 

2967 

2989 

2 

4 

7 

9 

11 

13 

16 

18 

20 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

2 

4 

6 

8 

11 

13 

15 

17 

19 

21 

3222 

3243 

3263 

3284 

3304 

3224 

3345 

3365 

3385 

3404 

2 

4 

6 

8 

10 

12 

14 

16 

18 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

2 

4 

6 

8 

10 

12 

14 

15 

17 

23 

3617 

3636 

3655 

3674 

3692 

3711 

3729 

3747 

3766 

3784 

2 

4 

6 

7 

9 

11 

13 

15 

17 

24 

3802 

3820 

3838 

3856 

3874 

3892 

3909 

3927 

3945 

3962 

2 

4 

5 

7 

9 

11 

12 

14 

16 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

2 

3 

5 

7 

9 

10 

12 

14 

15 

25 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

2 

3 

5 

7 

8 

10 

11 

13 

15 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

2 

3 

5 

6 

8 

9 

11 

13 

14 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

2 

3 

5 

6 

8 

9 

11 

12 

14 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

1 

3 

4 

6 

7 

9 

10 

12 

13 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

1 

3 

4 

6 

7 

9 

10 

11 

13 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

1 

3 

4 

6 

7 

8 

10 

11 

12 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

1 

3 

4 

5 

7 

8 

9 

11 

12 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

1 

3 

4 

5 

6 

8 

9 

10 

12 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

1 

3 

4 

5 

6 

8 

9 

10 

11 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

1 

2 

4 

5 

6 

7 

9 

10 

11 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

1 

2 

4 

5 

6 

7 

8 

10 

11 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

1 

2 

3 

5 

6 

7 

8 

9 

10 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

1 

2 

3 

5 

6 

7 

8 

9 

10 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

1 

2 

3 

4 

5 

7 

8 

9 

10 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

1 

2 

3 

4 

5 

6 

8 

9 

10 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

1 

2 

3 

4 

5 

6 

7 

8 

9 

42 

6232 

6243 

6253 

6263 

6274 

»‘.2M 

6294 

6304 

6314 

6325 

1 

2 

3 

4 

5 

6 

7 

8 

9 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

1 

2 

3 

4 

5 

6 

7 

8 

9 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

1 

2 

3 

4 

5 

6 

7 

8 

9 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

1 

2 

3 

4 

5 

6 

7 

8 

9 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

1 

2 

3 

4 

5 

6 

7 

7 

8 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

1 

2 

3 

4 

5 

5 

6 

7 

8 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

1 

2 

3 

4 

4 

5 

6 

7 

8 

49 

6902 

6911 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

1 

2 

3 

4 

4 

5 

6 

7 

8 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

1 

2 

3 

3 

4 

5 

6 

7 

8 

51 

7076 

7084 

7093 

7101 

7110 

7118 

7126 

7135 

7143 

7152 

1 

2 

3 

3 

4 

5 

6 

7 

8 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

1 

2 

2 

3 

4 

5 

6 

7 

7 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

1 

2 

2 

3 

4 

5 

6 

6 

7 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

1 

2 

2 

3 

4 

5 

6 

6 

7 











































APPENDIX 


169 


Four-Figure Logarithms. 


First two 
fL ures. 

Third figure. 

Fourth-figure 

differences. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1 

2 

3 

4 

5 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

1 

2 

2 

3 

4 

5 

5 

6 

7 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

1 

2 

2 

3 

4 

5 

5 

6 

7 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

1 

2 

2 

3 

4 

5 

5 

6 

7 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

7686 

7694 

7701 

1 

1 

2 

3 

4 

4 

5 

6 

7 

59 

7709 

7716 

7723 

7731 

7738 

7745 

7752 

7760 

7767 

7774 

1 

1 

2 

3 

4 

4 

5 

6 

7 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

1 

1 

2 

3 

4 

4 

5 

6 

6 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

1 

1 

2 

3 

4 

4 

5 

6 

6 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7980 

7987 

1 

1 

2 

3 

3 

4 

5 

6 

6 

63 

7993' 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

1 

1 

2 

3 

3 

4 

5 

5 

6 

64 

8062 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

1 

1 

2 

3 

3 

4 

5 

5 

6 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

1 

1 

2 

3 

3 

4 

5 

5 

6 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

1 

1 

2 

3 

3 

4 

5 

5 

6 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

1 

1 

2 

3 

3 

4 

5 

5 

6 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

1 

1 

2 

3 

3 

4 

4 

5 

6 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

1 

1 

2 

2 

3 

4 

4 

5 

6 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

1 

1 

2 

2 

3 

4 

4 

5 

6 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

1 

1 

2 

2 

3 

4 

4 

5 

5 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

1 

1 

2 

2 

3 

4 

4 

5 

5 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

1 

1 

2 

2 

3 

4 

4 

5 

5 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

1 

1 

2 

2 

3 

4 

4 

5 

5 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

1 

1 

2 

2 

3 

3 

4 

5 

5 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

1 

1 

2 

2 

3 

3 

4 

5 

5 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

1 

1 

2 

2 

3 

3 

4 

4 

5 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

1 

1 

2 

2 

3 

3 

4 

4 

5 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

1 

1 

2 

2 

3 

3 

4 

4 

5 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

1 

1 

2 

2 

3 

3 

4 

4 

5 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

1 

1 

2 

2 

3 

3 

4 

4 

5 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

1 

1 

2 

2 

3 

3 

4 

4 

5 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

1 

1 

2 

2 

3 

3 

4 

4 

5 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

1 

1 

2 

2 

3 

3 

4 

4 

5 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

1 

1 

2 

2 

3 

3 

4 

4 

5 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

1 

1 

2 

2 

3 

3 

4 

4 

5 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

0 

1 

1 

2 

2 

3 

3 

4 

4 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

0 

1 

1 

2 

2 

3 

3 

4 

4 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

0 

1 

1 

2 

2 

3 

3 

4 

4 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

0 

1 

1 

2 

2 

3 

3 

4 

4 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

0 

1 

1 

2 

2 

3 

3 

4 

4 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

0 

1 

1 

2 

2 

3 

3 

4 

4 

93 

9685 

9689 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

0 

1 

1 

2 

2 

3 

3 

4 

4 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

9768 

9773 

0 

1 

1 

2 

2 

3 

3 

4 

4 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

0 

1 

1 

2 

2 

3 

3 

4 

4 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

0 

1 

1 

2 

2 

3 

3 

4 

4 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

0 

1 

1 

2 

2 

3 

3 

4 

4 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

0 

1 

1 

2 

2 

3 

3 

4 

4 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

0 

1 

1 

2 

2 

3 

3 

3 

4 








































1T0 


CHEMICAL CALCULATIONS 


APPENDIX IV. — Antilogarithms. 


First two 
figures of 
mantissa. 

Third figure. 

Fourth-figure 

differences. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1 

2 

3 

4 5 

6 

7 

8 

9 

.00 

1000 

1002 

1005 

1007 

1009 

1012 

1014 

1016 

1019 

1021 

0 

0 

1 

1 

1 

1 

2 

2 

2 

.01 

1023 

1026 

1028 

1030 

1033 

1035 

1038 

1040 

1042 

1045 

0 

0 

1 

1 

1 

1 

2 

2 

2 

.02 

1047 

1050 

1052 

1054 

1057 

1059 

1062 

1064 

1067 

1069 

0 

0 

1 

1 

1 

1 

2 

2 

2 

.03 

1072 

1074 

1076 

1079 

1081 

1084 

1086 

1089 

1091 

1094 

0 

0 

1 

1 

1 

1 

2 

2 

2 

.04 

1096 

1099 

1102 

1104 

1107 

1109 

1112 

1114 

1117 

1119 

0 

1 

1 

1 

1 

2 

2 

2 

2 

.05 

1122 

1125 

1127 

1130 

1132 

1135 

1138 

1140 

1143 

1146 

0 

1 

1 

1 

1 

2 

2 

2 

2 

.06 

1148 

1151 

1153 

1156 

1159 

1161 

1164 

1167 

1169 

1172 

0 

1 

1 

1 

1 

2 

2 

2 

2 

.07 

1175 

1178 

1180 

1183 

1186 

1189 

1191 

1194 

1197 

1199 

0 

1 

1 

1 

1 

2 

2 

2 

2 

.08 

1202 

1205 

1208 

1211 

1213 

1216 

1219 

1222 

1225 

1227 

0 

1 

1 

1 

1 

2 

2 

2 

3 

.09 

1230 

1233 

1236 

1239 

1242 

1245 

1247 

1250 

1253 

1256 

0 

1 

1 

1 

1 

2 

2 

2 

3 

.10 

1259 

1262 

1265 

1268 

1271 

1274 

1276 

1279 

1282 

1285 

0 

1 

1 

1 

1 

2 

2 

2 

3 

.11 

1288 

1291 

1294 

1297 

1300 

1303 

1306 

1309 

1312 

1315 

0 

1 

1 

1 

2 

2 

2 

2 

3 

.12 

1318 

1321 

1324 

1327 

1330 

1334 

1337 

1340 

1343 

1346 

0 

1 

1 

1 

2 

2 

2 

2 

3 

.13 

1349 

1352 

1355 

1358 

1361 

1365 

1368 

1371 

1374 

1377 

0 

1 

1 

1 

2 

2 

2 

3 

3 

.14 

1380 

1384 

1387 

1390 

1393 

1396 

1400 

1403 

1406 

1409 

0 

1 

1 

1 

2 

2 

2 

3 

3 

.15 

1413 

1416 

1419 

1422 

1426 

1429 

1432 

1435 

1439 

1442 

0 

1 

1 

1 

2 

2 

2 

3 

3 

.16 

1445 

1449 

1452 

1455 

1459 

1462 

1466 

1469 

1472 

1476 

0 

] 

1 

1 

2 

2 

2 

3 

3 

.17 

1479 

1483 

1486 

1489 

1493 

1496 

1500 

1503 

1507 

1510 

0 

1 

1 

1 

2 

2 

2 

3 

3 

.18 

1514 

1517 

1521 

1524 

1528 

1531 

1535 

1538 

1542 

1545 

0 

1 

1 

1 

2 

2 

2 

3 

3 

.19 

1549 

1552 

1556 

1560 

1563 

1567 

1570 

1574 

1578 

1581 

0 

1 

1 

1 

2 

2 

3 

3 

3 

.20 

1585 

1589 

1592 

1596 

1600 

1603 

1607 

1611 

1614 

1618 

0 

1 

1 

1 

2 

2 

3 

3 

3 

.21 

1622 

1626 

1629 

1633 

1637 

1641 

1644 

1648 

1652 

1656 

0 

1 

1 

2 

2 

2 

3 

3 

3 

.22 

1660 

1663 

1667 

1671 

1675 

1679 

1683 

1687 

1690 

1694 

0 

1 

1 

2 

2 

2 

3 

3 

3 

.23 

1698 

1702 

1706 

1710 

1714 

1718 

1722 

1726 

1730 

1734 

0 

1 

1 

2 

2 

2 

3 

3 

4 

.24 

1738 

1742 

1746 

1750 

1754 

1758 

1762 

1766 

1770 

1774 

0 

1 

1 

2 

2 

2 

3 

3 

4 

.25 

1778 

1782 

1786 

1791 

1795 

1799 

1803 

1807 

1811 

1816 

0 

1 

1 

2 

2 

2 

3 

3 

4 

.28 

1820 

1824 

1828 

1832 

1837 

1841 

1845 

1849 

1854 

1858 

0 

1 

1 

2 

2 

3 

3 

3 

4 

.27 

1862 

1866 

1871 

1875 

1879 

1884 

1888 

1892 

1897 

1901 

0 

1 

1 

2 

2 

3 

3 

3 

4 

.28 

1905 

1910 

1914 

1919 

1923 

1928 

1932 

1936 

1941 

1945 

0 

1 

1 

2 

2 

3 

3 

4 

4 

.29 

1950 

1954 

1959 

1963 

1968 

1972 

1977 

1982 

1986 

1991 

0 

1 

1 

2 

2 

3 

3 

4 

4 

.30 

1995 

2000 

2004 

2009 

2014 

2018 

2023 

2028 

2032 

2037 

0 

1 

1 

2 

2 

3 

3 

4 

4 

.31 

2042 

2046 

2051 

2056 

2061 

2065 

2070 

2075 

2080 

2084 

0 

1 

1 

2 

2 

3 

3 

4 

4 

.32 

2089 

2094 

2099 

2104 

2109 

2113 

2118 

2123 

2128 

2133 

0 

1 

1 

2 

2 

3 

3 

4 

4 

.33 

2138 

2143 

2148 

2153 

215S 

2163 

2168 

2173 

2178 

2183 

0 

1 

1 

2 

2 

3 

3 

4 

4 

.34 

2188 

2193 

2198 

2203 

2208 

2213 

2218 

2223 

2228 

2234 

1 

1 

2 

2 

3 

3 

4 

4 

5 

.35 

2239 

2244 

2249 

2254 

2259 

2265 

2270 

2275 

2280 

2286 

1 

1 

2 

2 

3 

3 

4 

4 

5 

.36 

2291 

2296 

2301 

2307 

2312 

2317 

2323 

2328 

2333 

2339 

1 

1 

2 

2 

3 

3 

4 

4 

5 

.37 

2344 

2350 

2355 

2360 

2366 

2371 

2377 

2382 

2388 

2393 

1 

1 

2 

2 

3 

3 

4 

4 

5 

.38 

2399 

2404 

2410 

2415 

2421 

2427 

2432 

2438 

2443 

2449 

1 

1 

2 

2 

3 

3 

4 

4 

5 

.39 

2455 

2460 

2466 

2472 

2477 

2483 

2489 

2495 

2500 

2506 

1 

1 

2 

2 

3 

3 

4 

5 

5 

.40 

2512 

2518 

2523 

2529 

2535 

2541 

2547 

2553 

2559 

2564 

1 

1 

2 

2 

3 

4 

4 

5 

5 

.41 

2570 

2576 

2582 

2588 

2594 

2600 

2606 

2612 

2618 

2624 

1 

1 

2 

2 

3 

4 

4 

5 

5 

.42 

2630 

2636 

2642 

2649 

2655 

2661 

2667 

2673 

2679 

2685 

1 

1 

2 

2 

3 

4 

4 

5 

6 

.43 

2692 

2698 

2704 

2710 

2716 

2723 

2729 

2735 

2742 

2748 

1 

1 

2 

3 

3 

4 

4 

5 

6 

.44 

2754 

2761 

2767 

2773 

2780 

2786 

2793 

2799 

2805 

2812 

1 

1 

2 

3 

3 

4 

4 

5 

6 

.45 

2818 

2825 

2831 

2838 

2844 

2851 

2858 

2864 

2871 

2877 

1 

1 

2 

3 

3 

4 

5 

5 

6 

.46 

2884 

2891 

2897 

2904 

2911 

2917 

2924 

2931 

2938 

2944 

1 

1 

2 

3 

3 

4 

5 

5 

6 

.47 

2951 

2958 

2965 

2972 

2979 

2985 

2992 

2999 

3006 

3013 

1 

1 

2 

3 

3 

4 

5 

5 

6 

48 

3020 

3027 

3034 

3041 

3048 

3055 

3062 

3069 

3076 

3083 

1 

1 

2 

3 

4 

4 

5 

6 

6 

.49 

3090 

3097 

3105 

3112 

3119 

3126 

3133 

3141 

3148 

3155 

1 

1 

2 

3 

4 

4 

5 

6 

6 











































APPENDIX 


171 


Antilogarithms. 


First two 
figures of 
mantissa. 

Third figure. 

Fourth-figure 

differences. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.50 

3162 

3170 

8177 

3184 

3192 

3199 

3206 

3214 

3221 

3228 

1 

1 

2 

3 

4 

4 

5 

6 

7 

.51 

3236 

3243 

3251 

3258 

3266 

3273 

3281 

3289 

3296 

3304 

1 

2 

2 

3 

4 

5 

5 

6 

7 

.52 

3311 

3319 

3327 

3334 

3342 

3350 

3357 

3365 

3373 

3381 

1 

2 

2 

3 

4 

5 

5 

6 

7 

.53 

3388 

3396 

3404 

3412 

3420 

3428 

3436 

3443 

3451 

3459 

1 

2 

2 

3 

4 

5 

6 

6 

7 

.54 

3467 

3475 

3483 

3491 

3499 

3508 

3516 

3524 

3532 

3540 

1 

2 

2 

3 

4 

5 

6 

6 

7 

.55 

3548 

3556 

3565 

3573 

3581 

3589 

3597 

3606 

3614 

3622 

1 

2 

2 

3 

4 

5 

6 

7 

7 

.56 

3631 

3639 

3648 

3656 

3664 

3673 

3681 

3690 

3698 

3707 

1 

2 

3 

3 

4 

5 

6 

7 

8 

.57 

3715 

3724 

3733 

3741 

3750 

3758 

3767 

3776 

3784 

3793 

1 

2 

3 

3 

4 

5 

6 

7 

8 

.58 

3802 

3811 

3819 

3828 

3837 

3846 

3855 

3864 

3873 

3882 

1 

2 

3 

4 

4 

5 

6 

7 

8 

.59 

3890 

3899 

3908 

3917 

3926 

3936 

3945 

3954 

3963 

3972 

1 

2 

3 

4 

5 

5 

6 

7 

8 

.60 

3981 

3990 

3999 

4009 

4018 

4027 

4036 

4046 

4055 

4064 

1 

2 

3 

4 

5 

6 

6 

7 

8 

.61 

4074 

4083 

4093 

4102 

4111 

4121 

4130 

4140 

4150 

4159 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.62 

4169 

4178 

4188 

4198 

4207 

4217 

4227 

4236 

4246 

4256 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.63 

4266 

4276 

4285 

4295 

4305 

4315 

4325 

4335 

4345 

4355 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.64 

4365 

4375 

4385 

4395 

4406 

4416 

4426 

4436 

4446 

4457 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.65 

4467 

4477 

4487 

4498 

4508 

4519 

4529 

4539 

4550 

4560 

1 

2 

3 

4 

5 

6 

7 

8 

9 

.66 

4571 

4581 

4592 

4603 

4613 

4624 

4634 

4645 

4656 

4667 

1 

2 

3 

4 

5 

6 

7 

9 

10 

.67 

4677 

4688 

4699 

4710 

4721 

4732 

4742 

4753 

4764 

4775 

1 

2 

3 

4 

5 

7 

8 

9 

10 

.68 

4786 

4797 

4808 

4819 

4831 

4842 

4853 

4864 

4875 

4887 

1 

2 

3 

4 

6 

7 

8 

9 

10 

.69 

4898 

4909 

4920 

4932 

4943 

4955 

4966 

4977 

4989 

5000 

1 

2 

3 

5 

6 

7 

8 

9 10 

.70 

5012 

5023 

5035 

5047 

5058 

5070 

5082 

5093 

5105 

5117 

1 

2 

4 

5 

6 

7 

8 

9 

11 

.71 

5129 

5140 

5152 

5164 

5176 

5188 

5200 

5212 

5224 

5236 

1 

2 

4 

5 

6 

7 

8 

10 

11 

.72 

5248 

5260 

5272 

5284 

5297 

5309 

5321 

5333 

5346 

5358 

1 

2 

4 

5 

6 

7 

9 

10 

11 

.73 

5370 

5383 

5395 

5408 

5420 

5433 

5445 

5458 

5470 

5483 

1 

3 

4 

5 

6 

8 

9 

10 

11 

.74 

5495 j 

5508 

5521 

5534 

5546 

5559 

5572 

5585 

5598 

5610 

1 

3 

4 

5 

6 

8 

9 

10 

12 

.75 

5623 

5636 

5649 

5662 

5675 

5689 

5702 

5715 

5728 

5741 

1 

3 

4 

5 

7 

8 

9 10 12 

.76 

5754 

5768 

5781 

5794 

5S08 

5821 

5834 

5848 

5861 

5875 

1 

3 

4 

5 

7 

8 

9 

11 

12 

.77 

5888 

5902 

5916 

5929 

5943 

5957 

5970 

5984 

5998 

6012 

1 

3 

4 

5 

7 

8 

10 11 

12 

.78 

6026 

6039 

6053 

6067 

6081 

6095 

6109 

6124 

6138 

6152 

1 

3 

4 

6 

7 

8 

10 

11 

13 

.79 

6166 

6180 

6194 

6209 

6223 

6237 

6252 

6266 

6281 

6295 

1 

3 

4 

6 

7 

9 

10 11 

13 

.80 

6310 

6324 

6339 

6353 

6368 

6383 

6397 

6412 

6427 

6442 

1 

3 

4 

6 

7 

9 

10 12 

13 

.81 

6457 

6471 

6486 

6501 

6516 

6531 

6546 

6561 

6577 

6592 

2 

3 

5 

6 

8 

9 

11 

12 

14 

.82 

6607 

6622 

6637 

6653 

6668 

6683 

6699 

6714 

6730 

6745 

2 

3 

5 

6 

8 

9 

11 

12 

14 

.83 

6761 

6776 

6792 

6808 

6S23 

6839 

6855 

6871 

6887 

6902 

2 

3 

5 

6 

8 

9 

11 

13 14 

.84 

6918 

6934 

6950 

6966 

6982 

6998 

7015 

7031 

7047 

7063 

2 

3 

5 

6 

8 10 

11 

13 15 

.85 

7079 

7096 

7112 

7129 

7145 

7161 

7178 

7194 

7211 

7228 

2 

3 

5 

7 

8 10 

12 13 15 

.86 

7244 

7261 

7278 

7295 

7311 

7328 

7345 

7362 

7379 

7396 

2 

3 

5 

7 

8 

10 

12 

13 

15 

.87 

7413 

7430 

7447 

7464 

7482 

7499 

7516 

7534 

7551 

7568 

2 

3 

5 

7 

9 

10 

12 

14 

16 

.88 

75S6 

7603 

7621 

7638 

7656 

7674 

7691 

7709 

7727 

7745 

2 

4 

5 

7 

9 

11 

12 

14 

16 

.89 

7762 

7780 

7798 

7816 

7834 

7852 

7870 

7889 

7907 

7925 

2 

4 

5 

7 

9 

11 

13 14 16 

.90 

7943 

7962 

7980 

7998 

8017 

8035 

8054 

8072 

8091 

8110 

2 

4 

6 

7 

9 

11 

13 15 17 

.91 

8128 

8147 

8166 

8185 

8204 

8222 

8241 

8260 

8279 

8299 

2 

4 

6 

8 

9 

11 

13 

15 17 

.92 

8318 

8337 

8356 

8375 

8395 

8414 

8433 

8453 

8472 

8492 

2 

4 

6 

8 10 12 

14 

15 17 

.93 

8511 

8531 

8551 

8570 

8590 

8610 

8630 

8650 

8670 

8690 

2 

4 

6 

8 

10 12 

14 

16 18 

.94 

8710 

8730 

8750 

8770 

8790 

8810 

8831 

8851 

8872 

8892 

2 

4 

6 

8 10 12 

14 

16 18 

.95 

8913 

8933 

8954 

8974 

8995 

9016 

9036 

9057 

9078 

9099 

2 

4 

6 

8 10 12 

15 

17 

19 

.96 

9120 

9141 

9162 

9183 

9204 

9226 

9247 

9268 

9290 

9311 

2 

4 

6 

8 11 

13 

15 

17 

19 

.97 

9333 

9354 

9376 

9397 

9419 

9441 

9462 

9484 

9506 

9528 

2 

4 

7 

9 

11 

13 

15 17 20 

.98 

9550 

9572 

9594 

9616 

9638 

9661 

9683 

9705 

9727 

9750 

2 

4 

7 

9 

11 

13 

16 

18 20 

.99 

9772 

9795 

9817 

9840 

9863 

9886 

990S 

9931 

9954 

9977 

2 

5 

7 

9 11 

14 

16 18 20 
























































































































































































































































































































































INDEX 


PAGE 


Analysis, volumetric and 

gravimetric. 118 

Aqueous vapor, tension of.. . 28 

Atom, definition of. 35 

Atomic weight, definition of, 36, 46 
Avogadro’s hypothesis. 33 

Boiling-point. 29 

Boyle, law of. 9 

Charles, law of. 15 

Coefficient of expansion of 

gases. 15 

Cohesion in gases. 34 

Dalton, law of. 27 

Definite proportions, law of 46 

Density, absolute, definition 

of. 5 

relative, calculation of, 

from molecular weight 42 

definition of. 5 

standard of, for gases. . 6 

upon different standards 38 

relation of, to molecular 

weight. 34 

to pressure of gases.... 12 

to temperature of gases 18 

to pressure and tempera¬ 
ture of gases. 26 

Dyne, definition erf. 2 

Empirical formulae, definition 

of. 61 

derivation of, from percent¬ 
age composition. 60 

End-point in titrations. 106 

Erg, definition of . 3 

Equations, dependent. 92 

independent. 94 

study of. 82 

volume relations in. 86 


PAGE 


Force, unit of. 2 

Formulae, derivation of. 59 

for salts containing water 

of crystallization. ... 70 

from percentage composi¬ 
tion . 60,65 

from relation of formula- 
quantities to each 

other. 67 

from relation between for¬ 
mula-quantities and 
their corresponding 

weights. 71 

from relation between for¬ 
mula-quantities and 
percentage composi¬ 
tion. 74 

from single analytical val¬ 
ues . 75 

Formula-quantity, definition 

of. 48 

relation of, to molecular 

weight. 49 

to other formula-quan¬ 
tities.51,53 

to percentage composi¬ 
tion. 52 

Gay-Lussac, law of. 33 

Gram, definition of. 2 

Gram-molecular volume, defi¬ 
nition of. 40 

in chemical reactions. 122 

Gram-molecular weight, defi¬ 
nition of. 37 

Gravimetric analysis. 118 

Indicators in titrations. 106 

Iodimetry,. 158 

Ions, hydrogen and hydrox¬ 
ide . 106 


173 











































INDEX 


174 

PAGE 

Length, unit of. 1 

Liter, definition of. 2 

Mass, definition of. 3 

unit of. 2 

Meter, definition of. 1 

Molar solution, definition of 107 

Molecule, definition of. 33 

Molecular formulae, definition 

of. 46,61 

derivation of, from percent¬ 
age composition. 59 

Molecular volumes of gases, 

combinations between 122 
relation of, to volume-unit • 124 
Molecular weight, definition 


of. 34,46 

Neutralization. 106 

Normality factor. 109 

calculation of, by simple 

proportion. Ill 

Normal solution, definition of 108 

Oxidation, definition of. . . . 148 

Oxidizing solutions, with 

KjCrgOy. 150 

with KMn0 4 . 154 

Oxygen, as standard of rela¬ 
tive densities of gases. 6 

as standard of molecular 

weights. 6 

Partial pressures of gases ... 27 

Percentage composition of 

compounds. 46 

Percentage composition, re¬ 
lation of, to formula- 

quantities . 52 

as basis for percentage 

purity. 53 

Percentage purity. 53 

calculation of. 88 

Pressure, normal. 3 

relation of, to temperature 

of gases. 23 

to volume of gases. ... 11 

to volume and tempera¬ 
ture of gases. 26 

Pressure-f raction. 11 

Products resulting from mix¬ 
tures, calculation of.. 89 


PAGE 


Reaction-quantities. 82 

calculation of ratios be¬ 
tween .92,94 

calculation of, from weights 

of substances involved 97 

complex. 98 

Reduction, definition of .... 148 

Second, definition of. 1 

Solutions, adjustment of, to 

desired standard. 117 

comparison of, with stand¬ 
ard. 116 

standard. 108 

standard oxidizing.153,156 

Specific gravity, definition of 5 

Standardization of solutions, 

112,113 

Temperature, normal. 3 

relation of, to density of 

gases. 18 

to pressure of gases.... 23 

to volume of gases. 18 

to volume and pressure 

of gases. 25 

Temperature-fraction. 18 

Time, unit of. 1 

Titration of solutions.. 106 

Valence, definition of. 148 

Vapor density. 6 

Volume relations in chemical 

equations. 86 

Volume, relation of, to pres¬ 
sure of gases. 9 

to temperature of gases 17 

to temperature and pres¬ 
sure of gases.21, 24 

unit of. 2 

Volume-fraction. 11 

Volume-unit, for reactions 

between gases. 124 


calculation of, from known 
volumes of gases; 

125,126, 128 

calculation of, from ob¬ 
served volume-changes 129 
calculation of, from sev¬ 
eral simultaneous re¬ 


actions. .. 133,135 

Volumetric analysis......... 118 

Work, unit of. 3 
















































































































































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CHEMICAL BOOKS 


D. VAN NOSTRAND COMPANY 


American Institute of Chemical Engineers. Tansactions. 
Volume I., 1908 . folding plates. 8 vo. cloth. 212 pp. 

net, $ 6.00 

Annual Reports cn the Progress of Chemistry. Issued 
annually by the Chemical Society. 8 vo. cloth. Vol. 
I., 1904 , Vol. VL, 1909 , now ready, each, net, $2.00- 


BAILEY. R. 0. The Brewer's Analyst. Illustrated, 8 vo. 
cloth. 423 pp. net, $5.00 


BENNETT, HUGH G. The Manufacture of Leather. 

110 illustrations. 8 vo. cloth. 438 pp. net, $4.50 


BERNTHSEN, A. A Text-book of Organic Chemistry. 

English translation. Edited and revised by J. J. Sud- 
borough. Illustrated. 121 m cloth. 690 pp. 

net, $2.50 


BIRCHMORE, W. H. The Interpretation of Gas Analyses. 

Illustrated. 121110 . cloth. 75 PP- net, $1.25 



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D. VAN NO STRAND COMPANY’S 


BL3TTH, A. W. Foods: Their Composition and Analysis. 

A manual for the use of analytical chemists, with an 
introductory essay on the History of Adulterations. 
Sixth Edition, thoroughly revised, enlarged and re- 
zvritten. Illustrated. 8 vo. cloth. 634 pp. $7.50 


-Poisons: Their Effects and Detection. A manual for 

the use of analytical chemists and experts, with an 
introductory essay tm the Growth of Modern Toxicol¬ 
ogy. Fourth Edition, revised, enlarged and rewritten. 
Illustrated. 8 vo. cloth. 772 pp. $7.50 

BoCKMANN, F. Celluloid; Its Raw Material, Manufac¬ 
ture, Properties and Uses. 49 illustrations. i 2 mo. cloth. 
120 pp. net, $2.50 

BOOTH, WILLIAM H. Water Softening and Treatment. 

91 illustrations. 8 vo. cloth. 310 pp. net, $2.50 

B0URG0UGN0N, A. Physical Problems and Their Solu¬ 
tion. i 6 mo. boards. (Van Nostrand's Science Se¬ 
ries.) $0.50 

BRUCE, EDWIN M. Detection of the Common Food 
Adulterants. Illustrated. i 2 mo. cloth. 90 pp. 

net, $1.25 

BUSKETT, E. W. Fire Assaying. A practical treatise on 
the fire assaying of gold, silver and lead, including 
descriptions of the appliances used. Illustrated. i 2 mo. 
cloth. 112 pp. net, $1.25 




LIST OF CHEMICAL BOOKS 


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CAVEN, R. M., and LANDER, G. D. Systematic Inor¬ 
ganic Chemistry from the Standpoint of the Periodic 
Law. A text-book for advanced students. Illustrated. 
i 2 mo. cloth. 390 pp. net, $ 2.00 

CHURCH’S Laboratory Guide. A manual of practical 
chemistry for colleges and schools, specially arranged 
for agricultural students. Eighth Edition, revised and 
partly rewritten by Edward Kinch. Illustrated. 8 vo. 
cloth, 365 pp. net, $2.50 

CHRISTIE, W. W. Boiler-waters, Scale, Corrosion, Foam¬ 
ing. 77 illustrations. 8 vo. cloth. 242 pp. net, $3.00 

CORNWALL, H. B. Manual of Blow-pipe Analysis 

Qualitative and quantitative. With a complete system 
of determinative mineralogy. Sixth Edition, revised. 
70 illustrations. 8 vo. cloth. 310 pp. net, $2.50 

ECCLES, R. G. Food Preservatives; Their Advantages 
and Proper Use. The practical versus the theoretical 
side of the pure food problem. 8 vo. paper. 220 pp. 

$0.50 

ELIOT, C. W., and ST0RER, F. H. A Compendious Man¬ 
ual of Qualitative Chemical Analysis. Revised with 
the co-operation of the authors, by William R. 
Nichols. Twenty-second Edition, newly revised by 
W. B. Lindsay. Illustrated. i 2 mo. cloth. 205 pp. 

net, $1.25 

ENNIS, WILLIAM D. Linseed Oil and Other Seed Oils 

An industrial manual. 88 illustrations. 8 vo. cloth. 
336 pp. $4.00 



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D. VAN NO STRAND COMPANY'S 


FAY, I. W. Coal-tar Colors: Their Origin and Chemistry. 

Illustrated. 8 vo. cloth. In Press. 

FOYE, J. C. Chemical Problems. Fourth Edition , revised 
and enlarged. 161110 . cloth. 145 pp. (Van Nos¬ 
trand Science Series, No. 69 .) $0.50 

GROSSMANN, J. Ammonia and Its Compounds. ‘ Illus¬ 
trated. 121110 . cloth. 151 pp. net, $1.25 

HALE, WILLIAM J. Calculations in General Chemistry. 

With definitions, explanations and problems. Second 
Edition, revised. i 2 mo. cloth. 185 pp. net, $1.00 

HALL, CLARE H. Chemistry of Paints and Paint Ve¬ 
hicles. 8 vo. cloth. 141 pp. net, $ 2.00 

HOPKINS, N. M. Experimental Electrochemistry: Theo¬ 
retically and Practically Treated. 132 illustrations. 
8 vo. cloth. 298 pp. net, $3.00 

HOTJLLEVIGUE, L. The Evolution of the Sciences. 

8 vo cloth. 377 pp. net, $ 2.00 

INGLE, HERBERT. Manual of Agricultural Chemistry. 

Illustrated. 8 vo. cloth. 388 pp. net, $3.00 

JOHNSTON, J. F. W. Elements of Agricultural Chem¬ 
istry. Revised and lewritten by Charles A. Cameron 
and C. M. Aikman. Nineteenth Edition. Illustrated. 
i 2 mo. cloth. 502 pp. $2.60 


KEMBLE, W. F., and UNDERHILL, C. R. The Periodic 
Law and the Hydrogen Spectrum. Illustrated. 8 vo. 
paper. 16 pp. net. $0.50 



LIST OF CHEMICAL BOOKS 


5 


KEMP, J. F. A Handbook of Rocks. For use without 
the microscope. Fourth Edition revised. Illustrated. 
8 vo. cloth. 250 pp. net, $1.50 

KERSHAW, J. B. C. Fuel, Water, and Gas Analysis, for 
Steam Users. 50 illustrations. 8 vo. cloth. 178 pp. 

net, $2.50 

KOLLER, T. Cosmetics. A handbook of the manufac¬ 
ture, employment and testing of all cosmetic materials 
and cosmetic specialties. Translated from the German 
by Charles Salter. 8 vo. cloth. 262 pp. net, $2.50 


KRAUCH, C. Testing of Chemical Reagents for Purity. 
Authorized translation of the Third Edition, by J. A. 
Williamson and L. W. Dupre. With additions and 
emendations by the author. 8 vo. cloth. 350 pp. 

net, $3.00 

LASSAR-C0HN. Introduction to Modern Scientific 
. Chemistry. In the form of popular lectures suited for 
University Extension students and general readers. 
Translated from the Second German Edition by M. M. 
Pattison Muir. Illustrated. i 2 mo. cloth. 356 pp. 

$ 2.00 

LUNGE, GEORGE. Technical Methods of Chemical 
Analysis. Translated from the Second German Edition 
by Charles A. Keane, with the collaboration of eminent 
experts. To be complete in three volumes. Vol. I. (in 
two parts). 201 illustrations. 8 vo. cloth. 1024 pp. 

net, $15.00 

Volumes II. and III. in active preparation. 




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D. VAN NOSTRAND COMPANY’S 


LUNGE, GEORGE. Technical Chemists’ Handbook. 

Tables and methods of analysis for manufacturers of , 
inorganic chemical products. Illustrated. i 2 mo. 
leather. 276 pp. net, $3.50 

-Coal, Tar and Ammonia. Fourth and Enlarged Edi¬ 
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trations. 8 vo. cloth. 1210 pp. net, $15.00 

-The Manufacture of Sulphuric Acid and Alkali. 

A theoretical and practical treatise. 

Vol. I. Sulphuric Acid. Third Edition, enlarged. In 
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cloth. 1225 pp. net, $15.00 

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1044 pp. net, $15.00 

LUQUEUR, L. M. Minerals in Rock Sections. The 

practical methods of identifying them with the micro¬ 
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150 pp. net, $1!50 

MELICK, CHARLES W. Dairy Laboratory Guide. 52 

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MERCK, E. Chemical Reagents: Their Purity and Tests. 

8 vo. cloth. 250 pp. net, $1.50 

MILLER, E. H. Quantitative Analysis for Mining En¬ 
gineers. Second Edition. 8 vo. cloth. 158 pp. 

net, $1.50 





LIST OF CHEMICAL BOOKS 


7 


MOSES, A. J., and PARSONS, C. L. Elements of Mineral¬ 
ogy, Crystallography, and Blowpipe Analysis from a 
Practical Standpoint. Fourth Edition. 583 illustra¬ 
tions. 8 vo. cloth. 448 pp. net, $2.50 


MU&BY, A. E. Introduction to the Chemistry and 
Physics of Building Materials. Illustrated. 8 vo. cloth. 
365 pp. (Van Nostrand’s Westminster Series.) 

net, $ 2.00 

MURRAY, J. A. Soils and Manures. 33 illustrations. 
8 vo. cloth. 367 pp. (Van Nostrand’s Westminster 
Series.) net, $ 2.00 


NAQUET, A. Legal Chemistry. A guide to the detec¬ 
tion of poisons as applied to chemical jurisprudence. 
Translated, with additions, from the French, by J. P. 
Battershall. Second Edition , revised with additions. 
i 2 mo. cloth. 190 pp. $2.00 


OLSEN, J. C. A Textbook of Quantitative Chemical 

Analysis by Gravimetric and Gasoinetric Methods. 

Including 74 laboratory exercises giving the analysis 
of pure salts, alloys, minerals and technical products. 
Fourth Edition, revised and enlarged. 74 illustrations. 
8 vo. cloth., 576 pp. net, $4.00 

PARRY, ERNEST J. The Chemistry of Essential Oils 
and Artificial Perfumes. Second Edition , thoroughly 
revised and greatly enlarged. Illustrated. 8 vo. cloth. 
554 pp. net, $5.00 





8 


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PERKIN, F. M. Practical Methods of Inorganic Chem¬ 
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PHILLIPS. J. Engineering Chemistry. A pra-ctical 
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PLATTNER'S Manual of Qualitative and Quantitative 
Analysis with the Blowpipe. Eighth Edition, revised. 
Translated by Henry B. Cornwall, assisted by John 
H. Caswell, from the Sixth German Edition . by Fried¬ 
rich Kolbeck. 87 illustrations. 8vo. cloth. 463 pp. 

net, $ 4.00 

PRESCOTT, A. B. Organic Analysis. A manual of the 
descriptive and analytical chemistry of certain carbon 
compounds in common use. Sixth Edition. 8vo. cloth. 
533 PP- $ 5.00 

PRESCOTT, A. B., and JOHNSON, 0 . C. Qualitative 
Chemical Analysis. Sixth Edition, revised and en¬ 
larged. 8vo. cloth. 439 pp. net, $ 3.50 

PRESCOTT, A. B., and SULLIVAN, E. C. First Book in 
Qualitative Chemistry. For studies of water solution 
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PR 0 ST, E. Manual of Chemical Analysis. As applied 
to the assay of fuels, ores, metals, alloys, salts, and 
other mineral products. Translated from the original 
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net, $ 4.50 



LIST OF CHEMICAL BOOKS 


9 


PYNCHON, T. R. Introduction to Chemical Physics. 

Third Edition, revised and enlarged. 269 illustrations. 
8vo. cloth. 575 pp. $3.00 

ROGERS, ALLEN. A Laboratory Guide of Industrial 
Chemistry. Illustrated. 8vo. cloth. 170 pp. net, $ 1.50 

ROGERS, ALLEN, and ATJEERT, ALFRED B. Industrial 
Chemistry. Written by a staff of eminent specialists. 

In Press. 

ROTH, W. A. Exercises in Physical Chemistry. Author¬ 
ized translation by A. I. Cameron. 49 illustrations. 
8vo. cloth. 208 pp. net, $2.00 

SCHERER, R. Casein: Its Preparation and Technical 
Utilization. Translated from the German by Charles 
Salter. Illustrated. 8vo. cloth. 163 pp. net, $ 3.00 

SCHWEIZER, V. Distillation of Resins, Resinate Lakes 
and Pigments. 68 illustrations. 8vo. cloth. 183 pp. 

net, $ 3.50 

SEIDELL, A. Solubilities of Inorganic and Organic Sub¬ 
stances. A handbook of the most reliable quantitative 
solubility determinations. 121110. cloth. 367 pp. 

net, $ 3.00 

SEXTON, A. H. Fuel and Refractory Materials. Second 
Edition, revised. 104 illustrations. 121110. cloth. 
374 pp. net, $2.00 


SMITH, W. The Chemistry of Hat Manufacturing. 

Revised and edited by Albert Shonk. Illustrated. 
i2mo. cloth. 132 pp. net, $ 3.00 



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D. VAN NOSTRAND COMPANY’S 


SPEYERS, C. L. Text-book of Physical Chemistry. 20 

illustrations. 8vo. cloth. 230 pp. net, $2.25 

STEVENS, H. P. Paper Mill Chemist. 67 illustrations. 
82 tables. i6mo. cloth. 280 pp. net, $2.50 

SUDBOROUGH, J. J., and JAMES, J. C. Practical Or¬ 
ganic Chemistry. 92 illustrations. i2mo. cloth. 
394 pp. net, $2.00 

TITHERIEY, A. W. Laboratory Course of Organic 
Chemistry, Including Qualitative Organic Analysis. 

Illustrated. 8vo. cloth. 235 pp. net, $2.00 

r : w v .. " 

TOCH, M. Chemistry and Technology of Mixed Paints. 

62 photo-micrographs and engravings. 8vo. cloth. 

166 pp. net, $3.00 

TUCKER, J. H. A Manual of Sugar Analysis. Sixth 
Edition. 43 illustrations. 8vo. cloth. 353 pp. $3.50 

VAN NOSTRAND’S Chemical Annual, Based on Bieder- 

mann’s “Chemiker Kalender.” Edited by J. C. Olsen, 
with the co-operation of eminent chemists. Second 
Edition, revised and enlarged, 1909. i2mo. cloth. 

net, $2.50 

VINCENT, C. Ammonia and Its Compounds. Their 
manufacture and uses. Translated from the French 
by M. J. Salter. 32 illustrations. 8vo. cloth. 113 pp. 

net, $2.00 



LIST OF CHEMICAL BOOKS 


ii 


VON GEORGIEVICS, G. Chemical Technology of Textile 
Fibres. Translated from the German by Charles 
Salter. 47 illustrations. 8vo. cloth. 320 pp. net, $ 4.50 

-Chemistry of Dyestuffs. Translated from the Sec¬ 
ond German Edition by Charles Salter. 8vo. cloth. 

412 pp. net, $ 4.50 

WANKLYN, J. A. Milk Analysis. A practical treatise 
on the examination of milk and its derivatives, cream, 
butter and cheese. Illustrated. i2mo. cloth. 73 pp. 

$ 1.00 

—— Water Analysis. A practical treatise on the exami¬ 
nation of potable water. Eleventh Edition, revised, by 
W. J. Cooper. Illustrated. i2mo. cloth. 213 pp. 

$2.00 

WINCHELL, N. H. and A. N. Elements of Optical Min¬ 
eralogy. An introduction to microscopic petrography. 
350 illustrations. 4 plates. 8vo. cloth. 510 pp. 

$ 3.50 

WINKLER. C., and LUNGE, G. Handbook of Technical 
Gas Analysis. Second English Edition. Illustrated. 
8vo. cloth. 190 pp. $ 4.00 

WORDEN, C. E. The Nitrocellulose Industry. 275 illus¬ 
trations. 8vo. cloth. 800 pp. Two volumes. In Press. 

D. VAN NOSTRAND COMPANY 

Publishers and Booksellers 

33 nURRAY AND 37 WARREN STREETS, NEW YORK. 





SECOND ISSUE, 1909 

I2mo, cloth 575’,Pages Net, $2.50 


VAN NOSTRAND’S 

Chemical Annual 

A HANDBOOK OF USEFUL DATA 

for analytical, manufacturing and investi¬ 
gating chemists and chemical students. 

BASED ON BIEDERMANN’S “CHEMIKER KALENDER” 

EDITED BY 

Prof. J. C. OLSEN, A.M., Ph.D. 

Polytechnic Institute of Brooklyn ; formerly Fellow Johns Hopkins 
University ; author of “ Quantitative Chemical Analysis ” 

WITH THE CO-OPERATION OF EMINENT CHEMISTS 


CONTENTS 


New Tables—Physical and Chemical Constant ot' the Essential Oils and 
Alkaloids. 

Melting Points and Composition of the Fusible Alloys. 

Density of Carbon Dioxide Polenske Values of Oils, etc. 

Tables for the Calculation of Gravimetric, Volumetric, and Gas Analyses. 

Tables of the Solubility, Boiling and Freezing Points, Specific Gravity, and 
Molecular Weight of the commonly used Inorganic aud Organic Com¬ 
pounds. 

Specific Gravity Tables of Inorganic and Organic Compounds. 

Other Physical and Chemical Constants of Chemical and Technical Pro¬ 
ducts. 


Conversion Tables of Weights and Measures 
New Books and Current Literature of the Y 



ALL TABLF WILL BE REVISED ANNUALLY 'TO GIVE THE 
MOST RECENT AND ACCURATE DATA 


■a a. 


D. Van " Nostrand Company 

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